Tenth class state syllabus-text book-em-ap-ts-mathematics

Tenth class state syllabus-text book-em-ap-ts-mathematics

1. 1.1 INTRODUCTION
We have studied different types of numbers in earlier classes. We have learnt about natural
numbers, whole numbers, integers, rational numbers and irrational numbers. Let us recall a little
bit about rational numbers and irrational numbers.
Rational numbers are numbers which can be written in the form of
p
q
where both p and q
are integers and q is not equal to zero. They are a bigger set than integers as there can be many
rational numbers between two integers. All rational numbers can be written either in the form of
terminating decimals or non-terminating, repeating decimals.
Numbers which cannot be expressed in the form of
p
q
are irrational. These include numbers
like 2, 3, 5 and mathematical quantities like  etc. When these are written as decimals,
they are non-terminaing, non-recurring. For example, = 1.41421356... and  = 3.14159...
These numbers can be located on the number line.
The set of rational and irrational numbers together are called real numbers. We can show
them in the form of a diagram:
Rational Numbers
Integers
Whole Numbers
Natural
Numbers
Irrational
Numbers
Real Numbers
Real Numbers1

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In this chapter, we will see some theorems and the different ways in which we can prove them.
We will use the theorems to explore properties of rational and irrational numbers. Finally, we will
study about a type of function called logs (short for logarithmic) and see how they are useful in
science and everyday life.
But before exploring real numbers a little more, let us solve some questions.
EXERCISE - 1.1
1. Which of the following rational numbers are terminating and which are non-terminating,
repeating?
(i)
2
5
(ii)
17
18
(iii)
15
16
(iv)
7
40
(v)
9
11
2. Find any rational number between the pair of numbers given below:
(i)
1
2
and 1 (ii)
1
3
3
and
2
3
3
(iii)
4
9
and 2
3. Classify the numbers given below as rational or irrational.
(i)
1
2
2
(ii)24 (iii) 16 (iv) 7.7 (v)
4
9
(vii) 30 (viii) 81
4. Show the following real numbers on the number line. If necessary make a seperate
number line for each number.
(i)
3
4
(ii)
9
10

(iii)
27
3
(iv) 5 (v) 16
THINK AND DISCUSS
Are all integers also included in real numbers? Why?
1.2 EXPLORING REAL NUMBERS
Let us explore real numbers more in this section. We know that natural numbers are also
included in real numbers. So, we will start with them.
1.2.1 THE FUNDAMENT AL THEOREM OF ARITHMETIC
In earlier classes, we have seen that all natural numbers, except 1, can be written as a
product of their prime factors. For example, 3 = 3, 6 as 2  3, 253 as 11  23 and so on.
(Remember: 1 is neither a composite nor a prime). Do you think that there may be a composite

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number which is not the product of the powers of primes? To answer this, let us factorize a
natural number as an example.
We are going to use the factor tree which you all are familiar with. Let us take some large
number, say 163800, and factorize it as shown :
So we have factorized 163800 as 2  2  2  3  3  5  5  7  13. So 163800
= 2
3
 3
2
 5
2
 7  13, when we write it as a product of power of primes.
Try another number, 123456789. This can be written as 3
2
 3803  3607. Of course,
you have to check that 3803 and 3607 are primes! (Try it out for several other natural numbers
yourself.) This leads us to a conjecture that every composite number can be written as the
product of powers of primes.
Now, let us try and look at natural numbers from the other direction. Let us take any
collection of prime numbers, say 2, 3, 7, 11 and 23. If we multiply some or all of these numbers,
allowing them to repeat as many times as we wish, we can produce a large collection of positive
integers (in fact, infinitely many). Let us list a few :
2  3  11 = 66 7  11 = 77

163800
2 81900
2 40950
2 20475
3 6825
3 2275
5 455
5 91
7 13

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7  11  23 = 1771 3  7  11 1  23 = 5313
2  3  7  11 1  23 = 10626 2
3
 3  7
3
= 8232
2
2
 3  7  11 1  23 = 21252
Now, let us suppose your collection of primes includes all the possible primes. What is
your guess about the size of this collection? Does it contain only a finite number of composites or
infinitely many? In fact, there are infinitely many primes. So, if we multiply all these primes in all
possible ways, we will get an infinite collection of composite numbers.
This gives us the Fundamental Theorem of Arithmetic which says that every composite
number can be factorized as a product of primes. Actually, it says more. It says that given any
composite number it can be factorized as a product of prime numbers in a ‘unique’ way, except
for the order in which the primes occur. For example, when we factorize 210, we regard 2  3
 5  7 as same as 3  5  7  2, or any other possible order in which these primes are
written. That is, given any composite number there is one and only one way to write it as a
product of primes, as long as we are not particular about the order in which the primes occur. Let
us now formally state this theorem.
Theorem-1.1 : (Fundamental Theorem of Arithmetic) : Every composite number can be
expressed (factorised) as a product of primes, and this factorization is unique, apart
from the order in which the prime factors occur.
In general, given a composite number x, we factorize it as x = p
1
p
2
...p
n
, where p
1
, p
2
...,
p
n
are primes and written in ascending order, i.e., p
1
 p
2
... p
n
. If we use the same primes,
we will get powers of primes. Once we have decided that the order will be ascending, then the
way the number is factorised, is unique. For example,
163800 = 2  2  2  3  3  5  5  7  13 = 2
3
 3
2
 5
2
 7  13
TRY THIS
Write 2310 it as a product of prime factors. Also see how 2-3 of your friends have
factorized the number. Have they done it like you? Verify your final product with your friend’s
result. Try this for 3 or 4 more numbers. What do you conclude?
While this is a result that is easy to state and understand, it has some very deep and
significant applications in the field of mathematics. Let us see two examples.
You have already learnt how to find the HCF (Highest Common Factor) and LCM
(Lowest Common Multiple) of two positive integers using the Fundamental Theorem of Arithmetic

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in earlier classes, without realizing it! This method is also called the prime factorization method.
Let us recall this method through the following example.
Example-1. Find the LCM and HCF of 12 and 18 by the prime factorization method.
Solution : We have12 = 2  2  3 = 2
2
 3
1
18 = 2  3  3 = 2
1
 3
2
Note that HCF (12, 18) = 2
1
 3
1
= 6=Product of the smallest power of each
common prime factors in the numbers.
LCM (12, 18) = 2
2
 3
2
= 36 =Product of the greatest power of each
prime factors, in the numbers.
From the example above, you might have noticed that HCF (12, 18)  LCM (12, 18)
= 12  18. In fact, we can verify that for any two positive integers a and b, HCF (a,b)  LCM
(a, b) = a  b. We can use this result to find the LCM of two positive integers, if we have
already found the HCF of the two positive integers.
Example 2. Consider the numbers 4
n
, where n is a natural number. Check whether there is any
value of n for which 4
n
ends with the digit zero?
Solution : For the number 4
n
to end with digit zero for any natural number n, it should be
divisible by 5. This means that the prime factorisation of 4
n
should contain the prime number 5.
But it is not possible because 4
n
= (2)
2n
so 2 is the only prime in the factorisation of 4
n
. Since 5
is not present in the prime factorization, so there is no natural number n for which 4
n
ends with the
digit zero.
TRY THIS
Show that 12
n
cannot end with the digit 0 or 5 for any natural number ‘n’.
EXERCISE - 1.2
1.Express each number as a product of its prime factors.
(i) 140 (ii) 156 (iii) 3825(iv) 5005(v) 7429
2.Find the LCM and HCF of the following integers by applying the prime factorization
method.
(i) 12, 15 and 21(ii) 17, 23, and 29(iii) 8, 9 and 25
(iv) 72 and 108(v) 306 and 657

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3.Check whether 6
n
can end with the digit 0 for any natural number n.
4.Explain why 7  11  13 + 13 and 7  6  5  4  3  2  1 + 5 are composite
numbers.
5.How will you show that (17  11  2) + (17  11 1  5) is a composite number?
Explain.
Now, let us use the Fundamental Theorem of Arithmetic to explore real numbers further.
First, we apply this theorem to find out when the decimal expansion of a rational number is
terminating and when it is non-terminating, repeating. Second, we use it to prove the irrationality
of many numbers such as 2, 3and 5.
1 . 2 . 2RATIONAL NUMBERS AND THEIR DECIMAL EXPANSIONS
In this section, we are going to explore when their decimal expansions of rational numbers
are terminating and when they are non-terminating, repeating.
Let us consider the following terminating decimal expressions of some rational numbers:
(i) 0.375(ii) 1.04(iii) 0.0875(iv) 12.5(v) 0.00025
Now let us express them in the form of
p
q
.
(i)
3
375 375
0.375
1000
10
 
(ii)
2
104 104
1.04
100 10
 
(iii)
4
875 875
0.0875
10000 10
  (iv)
1
125 125
12.5
10 10
 
(v) 0.00025 =
25
100000
=
5
25
10
We see that all terminating decimals taken by us can be expressed as rational numbers
whose denominators are powers of 10. Let us now prime factorize the numerator and denominator
and then express in the simplest rational form :
Now(i)
3
3 3 3 3
375 3 5 3 3
0.375
10 2 5 2 8

   


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(ii)
3
2 2 2 2
104 2 13 26 26
1.04
10 2 5 5 25

   

(iii)
3
4 4 4 4
875 5 7 7
0.0875
10 2 5 2 5

  
 
(iv)
3
125 5 25
12.5
10 2 5 2
  

(v)
2
5 5 5 5 3
25 5 1 1
0.00025
10 2 5 2 5 4000
   
 
Do you see a pattern in the denominators? It appears that when the decimal expression
is expressed in its simplest rational form then p and q are coprime and the denominator (i.e., q)
has only powers of 2, or powers of 5, or both. This is because the powers of 10 can only have
powers of 2 and 5 as factors.
DO THIS
Write the following terminating decimals as rational numbers in the form of
p
q
, q0 and
p, q are co-primes
(i) 15.265(ii) 0.1255(iii) 0.4(iv) 23.34 (v) 1215.8
What can you conclude about the denominators through this process?
LET US CONCLUDE
Even though, we have worked only with a few examples, you can see that any rational
number which has a decimal expansion that terminates can be expressed as a rational number
whose denominator is a power of 10. The only prime factors of 10 are 2 and 5. So, when we
simplyfy the rational number, we find that the number is of the form
p
q
, where the prime
factorization of q is of the form 2
n
5
m
, and n, m are some non-negative integers.
We can write our result formally :
Theorem-1.2 : Let x be a rational number whose decimal expansion terminates. Then x
can be expressed in the form
p
q
, where p and q are coprime, and the prime factorization
of q is of the form 2
n
5
m
, where n, m are non-negative integers.

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You are probably wondering what happens the other way round. That is, if we have a
rational number of the form
p
q
, and the prime factorization of q is of the form 2
n
5
m
, where n, m
are non-negative integers, then does
p
q
have a terminating decimal expansion?
So, it seems to make sense to convert a rational number of the form
p
q
, where q is of the
form 2
n
5
m
, to an equivalent rational number of the form
a
b
, where b is a power of 10. Let us go
back to our examples above and work backwards.
(i)
3
25 5 125
12.5
2 2 5 10
  

(ii)
3
2 2 2 2
26 26 13 2 104
1.04
25 5 2 5 10

   

(iii)
3
3 3 3 3
3 3 3 5 325
0.375
8 2 2 5 10

   

(iv)
3
4 4 4 4
7 7 7 5 875
0.0875
80 2 5 2 5 10

   
 
(v)
2
5 3 5 5 5
1 1 5 25
0.00025
4000 2 5 2 5 10
   
 
So, these examples show us how we can convert a rational number of the form
p
q
,
where q is of the form 2
n
5
m
, to an equivalent rational number of the form
a
b
, where b is a power
of 10. Therefore, the decimal expansion of such a rational number terminates. We find that a
rational number of the form
p
q
, where q is a power of 10, will have terminating decimal expansion.
So, we find that the converse of theorem 1 is alo true and can be formally stated as :
Theorem 1.3 : Let x =
p
q
be a rational number, such that the prime factorization of q is
of the form 2
n
5
m
, where n, m are non-negative integers. Then x has a decimal expansion
which terminates.

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DO THIS
Write the following rational numbers of the form
p
q
, where q is of the form 2
n
5
m
where
n, m are non-negative integers and then write the numbers in their decimal form
(i)
3
4
(ii)
7
25
(iii)
51
64
(iv)
14
23
(v)
80
81
1.2.3 NON-TERM INATING, RECURRING RATIONAL NUMBERS
Let us now consider rational numbers whose decimal expansions are
non-terminating and recurring. Once again, let us look at an example to see
what is going on-
Let us look at the decimal conversion of
1
7
.
1
7
= 0.1428571428571 ..... which is a non-terminating and recurring
decimal. Notice, the block of digits '142857' is repeating in the quotient.
Notice that the denominator here, i.e., 7 is not of the form 2
n
5
m
.
DO THIS
Write the following rational numbers as decimals and find out the
block of digits, repeating in the quotient.
(i)
1
3
(ii)
2
7
(iii)
5
11
(iv)
10
13
From the 'do this exercise' and from the example taken above, we can formally state:
Theorem-1.4 : Let x =
p
q
be a rational number, such that the prime factorization of q is
not of the form 2
n
5
m
, where n, m are non-negative integers. Then, x has a decimal
expansion which is non-terminating repeating (recurring).
From the discussion above, we can conclude that the decimal expansion of every rational
number is either terminating or non-terminating repeating.
0.1428571
7 1.0000000
7
30
28
20
14
60
56
40
35
50
49
10
7
30

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Example-3. Using the above theorems, without actual division, state whether the following
rational numbers are terminating or non-terminating, repeating decimals.
(i)
16
125
(ii)
25
32
(iii)
100
81
(iv)
41
75
Solution :(i)
3
16 16 16
125 5 5 5 5
 
 
= terminating decimal.
(ii)
5
25 25 25
32 2 2 2 2 2 2
 
   
= terminating decimal.
(iii)
4
100 100 10
81 3 3 3 3 3
 
  
= non-terminating, repeating decimal.
(iv)
2
41 41 41
75 3 5 5 3 5
 
  
= non-terminating, repeating decimal.
Example-4. Write the decimal expansion of the following rational numbers without actual division.
(i)
35
50
(ii)
21
25
(iii)
7
8
Solution :(i)
1
35 7 5 7 7
0.7
50 2 5 5 2 5 10

   
  
(ii)
2
2 2 2 2
21 21 21 2 21 4 84
0.84
25 5 5 5 5 2 5 2 10
 
    
   
(iii)
 
3
3 33 3 3
7 7 7 7 5 7 25 875
0.875
8 2 2 2 2 2 5 2 5 10
 
     
   
EXERCISE - 1.3
1.Write the following rational numbers in their decimal form and also state which are
terminating and which have non-terminating, repeating decimal expansions.
(i)
3
8
(ii)
229
400
(iii)
1
4
5
(iv)
2
11
(v)
8
125
2.Without actually performing long division, state whether the following rational numbers
will have a terminating decimal expansion or a non-terminating, repeating decimal
expansion.

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(i)
13
3125
(ii)
11
12
(iii)
64
455
(iv)
15
1600
(v)
29
343
(vi)
3 2
23
2 5
(vii)
2 7 5
129
2 5 7
(viii)
9
15
(ix)
36
100
(x)
77
210
3.Write the decimal expansion of the following rationals using Theorem 1.1.
(i)
13
25
(ii)
15
16
(iii)
3 2
23
2 .5
(iv)
2 2
7218
3 .5
(v)
143
110
4.The decimal expansions of some real numbers are given below. In each case, decide
whether the number is rational or not. If it is rational, and of the form
p
q
, what can you
say about the prime factors of q?
(i) 43.123456789(ii) 0.120120012000120000… (iii) 43.123456789
1.3 MORE ABOUT IRRATIONAL NUMBERS
Recall, a real number ‘S’ is called irrational if it cannot be written in the form
p
q
, where
p and q are integers and q0. Some examples of irrational numbers, with which you are already
familiar, are :
2
2, 3, 15, , ,
3
  0.10110111011110…, etc.
In this section, we will prove irrational numbers to be irrational with the help of the
fundamental theorem of arithmetic. We will prove that 2, 3, 5and in general, p is
irrational, where p is a prime.
Before we prove that 2 is irrational, we will look at a statement, the proof of which is
based on the Fundamental Theorem of Arithmetic.
Statement-1 : Let p be a prime number. If p divides a
2
, where a is a positive integer, then
p divides a.

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Proof : Let a be any positive integer. Then the prime factorization of a is as follows :
a = p
1
p
2
… p
n
, where p
1
, p
2
, …., p
n
are primes, not necessarily distinct.
Therefore a
2
= (p
1
p
2
… p
n
) (p
1
p
2
… p
n
) = p
2
1
p
2
2
… p
2
n
.
Now, here we have been given that p divides a
2
. Therefore, from the Fundamental
Theorem of Arithmetic, it follows that p is one of the prime factors of a
2
. Also, using the uniqueness
part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a
2
are p
1
p
2
… p
n
. So p is one of p
1,
p
2,
… p
n
.
Now, since p is one of p
1
p
2
… p
n
, p therefore, it divides a.
DO THIS
Verify the statement proved above for p= 2 and p = 5 and for a
2
= 1, 4, 9, 25, 36, 49, 64
and 81.
We are now ready to give a proof that 2 is irrational. We will use a technique called
proof by contradiction.
Example-5. Prove that 2 is irrational.
Proof : Since we are using proof by contradiction, let us assume the contrary, i.e., 2 is rational.
If it is rational, then there must exist two integers r and s (s0) such that 2 =
r
s
.
Suppose r and s have a common factor other than 1. Then, we divide by the common
factor to get 2 =
a
b
, where a and b are co-prime.
So, b2= a.
Squaring on both sides and rearranging, we get 2b
2
= a
2
. Therefore, 2 divides a
2
.
Now, by statement 1, it follows that if 2 divides a
2
it also divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b
2
= 4c
2
, that is, b
2
= 2c
2
.
This means that 2 divides b
2
, and so 2 divides b (again using statement 1 with p= 2).
Therefore, both a and b have 2 as a common factor.

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But this contradicts the fact that a and b are co-prime and have no common factors other
than 1.
This contradiction has arisen because of our assumption that 2 is rational. So, we
conclude that 2 is irrational.
In general, it can be shown that d is irrational whenever d is a positive integer which is
not the square of an integer. As such, it follows that 6, 8, 15 24 etc. are all irrational
numbers.
In earlier classes, we mentioned that :
• the sum or difference of a rational and an irrational number is irrational and
• the product and quotient of a non-zero rational and irrational number is irrational.
We prove some particular cases here.
Example-6. Show that 5 – 3 is irrational.
Solution : Let us assume, to the contrary, that 5 – 3 is rational.
That is, we can find coprimes a and b (b 0) such that 5 – 3 =
a
b
.
Therefore, 5 –
a
b
= 3
Rearranging this equation, we get 3 = 5
5a b a
b b

 
Since a and b are integers, we get 5
a
b
 is rational so 3 is rational.
But this contradicts the fact that 3 is irrational.
This contradiction has arisen because of our incorrect assumption that 5 – 3 is rational.
So, we conclude that 5 – 3 is irrational.
Example-7. Show that 32 is irrational.
Solution : Let us assume, the contrary, that 32 is rational.
i.e., we can find co-primes a and b (b 0) such that 32 =
a
b
.

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Rearranging, we get 2=
3
a
b
.
Since 3, a and b are integers,
3
a
b
is rational, and so 2 is irrational.
But this contradicts the fact that 2 is irrational.
So, we conclude that 3 2 is irrational.
Example-8. Prove that 2 + 3 is irrational.
Solution : Let us suppose that 2 + 3 is irrational.
Let 2 + 3 =
a
b
, where
a
b
is a rational, b  0
Therefore, 2=
a
b
–3 .
Squarring on both sides, we get
2
2
2 3 2 3
a a
b b
  
Rearranging

2
2
2
3 3 2
a a
b b
  

2
2
5
a
b


2 2
5
3
2
a b
ab


Since a, b are integers,
2 2
5
2
a b
ab

is rational, and so, 3 is rational.
This contradicts the fact that 3 is irrational. Hence, 2 3 is irrational.
Note :
1. The sum of the two irrational numbers need not to be irrational.
For example, if a = 2 and b =2, then both a and b are irrational, but a + b = 0
which is rational.

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15
2. The product of two irrational numbers need not to be irrational.
For example, a =2and b =8, then both a and b are irrational, but
ab = 16 = 4 which is rational.
EXERCISE - 1.4
1. Prove that the following are irrational.
(i)
1
2
(ii) 3 + 5 (iii) 6 + 2(iv) 5 (v) 3 + 25
2. Prove that p q is irrational, where p, q are primes.
TRY THIS
Properties of real numbers
In this chapter, you have seen many examples to show whether a number is rational or
irrational. Now assuming that a, b and c represent real numbers, use your new knowledge to find
out whether all the properties listed below hold for real numbers. Do they hold for the operations
of subtraction and division? Take as many real numbers you want and investigate.
Property Addition Multiplication
1. Closure a + b = c a . b = c
2. Commutative a + b = b + a a . b = b.a
3. Associative a + (b + c) = (a + b) + ca(bc) = (ab).c
4. Identity a + 0 = 0 + a = a a.1 = 1.a = a
5. Inverse a + (-a) = 0 a.
1
a
= 1, (a0)
6. Distributive a (b + c) = ab + ac
1.5 UNDERSTANDING LOGARITHMS
In this section, we are going to learn about logarithms. Logarithms are used for all sorts
of calculations in engineering, science, business, economics and include calcuating compound
interest, exponential growth and decay, pH value in chemistry, measurement of the magnitude of
earthquakes etc.

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16
However, before we can deal with logarithms, we need to revise the laws of exponents
as logarithms and laws of exponents are closely related.
1.5.1 EX PO NENTS REVISTED
We know that when 81 is written as 3
4
it is said to be written in its exponential form. That
is, in 81 = 3
4
, the number 4 is the exponent or index and 3 is the base. We say that -
81 is the 4
th
power of the base 3 or 81 is the 4
th
power of 3. Similarly, 27 = 3
3
.
Now, suppose we want to multiply 27 and 81; one way of doing this is by directly
multiplying. But multiplication could get long and tedious if the numbers were much larger than 81
and 27. Can we use powers to makes our work easier?
We know that 81 = 3
4
. We also know that 27 = 3
3
.
Using the Law of exponents
n m m n
a a a

  , we can write
27 x 81 = 3
3
 3
4
= 3
7
Now, if we had a table containing the values for the powers of 3, it would be straight
forward task to find the value of 3
7
and obtain the result of 81  27 = 2187.
Similaly, if we want to divide 81 by 27 we can use the laws of exponents
m n m n
a a a

  where m > n. Then, 81 ÷ 27 = 3
4
÷ 3
3
= 3
1
or simply 3
Notice that by using powers, we have changed a multiplication problem into one involving
addition and a division problem into one of subtration i.e., the addition of powers, 4 and 3 and
the subtraction of the powers 4 and 3.
DO THIS
Try to write the numbers 10, 100, 1000, 10000 and 100000 in exponential forms.
Identify the base and index in each case.
TRY THESE
Find 16 x 64, without actual multiplication, using exponents.
Find 25 x 125, without actual multiplication, using exponents.
Express 128 and 32 as powers of 2 and find 128 ÷ 32.

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17
1.5.2 WRITING EX PONENTS AS LOGARITHMS
We know that 10000 =10
4
. Here, 10 is the base and 4 is the exponent. Writing a
number in the form of a base raised to a power is known as exponentiation. We can also write
this in another way called logarithms.
For example, we write log
10
10000 = 4. This is stated as “log base 10 of 10000 equals
4”.
We observe that the base in the original expression becomes the base of the logarithm.
Thus,
10000=10
4
is the same as log
10
10000 = 4.
In general, we write log
a
x = n if a
n
= x; where a and x are positive numbers and
a  1. Let us understand this better through examples.
Example-9. Write i) 64 = 8
2
ii) 64 = 4
3
Solution : (i) The logarithmic form of 64 = 8
2
is log
8
64 = 2.
(ii) The logarithmic form of 64 = 4
3
is log
4
64 = 3.
In this example, we find that log base 8 of 64 is 2 and log base 4 of 64 is 3. So, the
logarithms of the same number to different bases are different.
DO THIS
Write 16 = 2
4
in logarithmic form. Is it the same as log
4
16?
Example-10. Write the exponential form of the following .
(i) log
10
100 = 2 (ii) log
5
25 = 2 (iii) log
2
2 = 1 (iv) log
10
10 = 1
Solution :(i) Exponential form of log
10
100=2 is 10
2
= 100.
(ii) Exponential form of log
5
25=2 is 5
2
= 25.
(iii) Exponential form of log
2
2=1 is 2
1
= 2.
(iv) Exponential form of log
10
10=1 is 10
1
= 10.
In cases (iii) and (iv), we notice that log
10
10 =1 and log
2
2=1. In general, for any base a,
a
1
= a, log
a
a=1. a
1
= a so log
a
a = 1
TRY THIS
Show that a
0
= 1 so log
a
1=0.

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18
DO THESE
1. Write the following in logarithmic form.
(i) 11
2
= 121 (ii) (0.1)
2
= 0.01(iii) a
x
= b
2. Write the following in exponential form.
(i) log
5
125 = 3 (ii) log
4
64 = 3 (iii) log
a
x = b (iv) log
2
2 = 0
Example-11. Determine the value of the following logarithms.
(i) log
3
9 (ii) log
8
2 (iii) log
c
c
Solution : (i) Let log
3
9=x, then the exponential form is 3
x
=9  3
x
=3
2
 x=2
(ii) Let log
8
2=y, then the exponential form is 8
y
=2  (2
3
)
y
=2  3y=1  y =
1
3
(iii) Let log
c
c=z, then the exponential form is
1
2
1
2
z z
c c c c z    
1.5.3 LAWS OF ALGORITHM S
Just like we have rules or laws of exponents, we have three laws of logarithms. We will
try to prove them in the coming sections
1.5.3a The first law of logarithms
Suppose x = a
n
and y = a
m
where a>0 and a1. Then we know that we can write:
log
a
x = n andlog
a
y = m .............. (1)
Using the first law of exponents we know thata
n
 a
m
= a
n+m
So,xy = a
n
 a
m
= a
n+m
i.e.xy = a
n+m
Writing in the logarithmic form, we get
log
a
xy = n+m .............. (2)
But from 1, n = log
a
x and m=log
a
y.
So, log
a
xy = log
a
x + log
a
y
So, if we want to multiply two numbers and find the logarithm of the product, we can do
this by adding the logarithms of the two numbers. This is the first law of logarithms.
log
a
xy = log
a
x + log
a
y

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19
1.5.3b The second law of logarithms states log
a

x
y
= log
a
x - log
a
y
TRY THIS
Prove the second law of logarithms by using the law of exponents
n
n m
m
a
a
a


1.5.3c The third law of logarithms
Let x = a
n
so log
a
x = n. Suppose, we raise both sides of x = a
n
to the power m, we get-
x
m
= (a
n
)
m
Using the laws of exponents-
x
m
= a
nm
If we think of x
m
as a single quantity, the logarithmic form of it, is
log
a
x
m
= nm
log
a
x
m
= m log
a
x (a
n
= x so log
a
x
= n)
This is the third law. It states that the logarithm of a power of a number can be obtained
by multiplying the logarithm of the number by that power.
log
a
x
m
= m log
a
x
Example-12. Expand log15
Solution : As you know, log
a
xy = log
a
x + log
a
y.
So, log15= log (3 × 5)
= log3 + log5
Example-13. Expand log
343
125
Solution : As you know, log
a

x
y
= log
a
x - log
a
y
So, log
343
125
= log343 – log125
= log7
3
– log5
3

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20
Since, log
a
x
m
= m log
a
x
= 3log7 – 3log5
So log
343
125
= 3(log7 – log5).
Example-14. Write 2log3 + 3log5 – 5log2 as a single logarithm.
Solution :2log3 + 3log5 – 5log2
= log3
2
+ log5
3
– log2
5
(since in m log
a
x=log
a
x
m
)
= log9 + log125 – log32
= log (9 × 125) – log32 (Since log
a
x + log
a
y = log
a
xy)
= log1125 – log32
= log
1125
32
(Since log
a
x – log
a
y = log
a
x
y
)
DO THESE
1. Write the following in the form log
a
x + log
a
y
(i)8  32 (ii)49  343(iii)81  729
2. Write the following in the form log
a
x - log
a
y
(i)8 ÷ 64 (ii)81 ÷ 27
3. Write the following in logarithmic forms
(i)4
3
= (2
2
)
3
(ii)36
2
= (6
2
)
2
EXERCISE - 1.5
1.Write the following using logarithms.
(i)3
5
= 243 (ii)2
10
= 1024(iii)10
6
= 1000000
(iv)10
-3
= 0.001(v)3
-2
=
1
9
(vi)6
0
= 1
(vii)5
-1
=
1
5
(viii)49 7 (ix)
2
3
27 9
(x)
2
5
1
32
4



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21
2.Write the following using powers.
(i)log
18
324 = 2(ii)log
10
10000 = 4 (iii)log
ax= b
(iv)
8
4
logx (v)log
3
1
27
 

 
= y
3.Determine the value of the following.
(i)log
25
5 (ii)log
81
3 (iii)log
2
1
16
 

 
(iv)log
7
1 (v)log
xx (vi)log
2
512
(vii)log
10
0.01 (viii) 2
3
8
log
27
 

 
4.Simplify each of the following expressions as log N. Determine the value of N. (You can
assume the base is 10, but the results are identical which ever base is used).
(i)log 2 + log 5 (ii)log 16 - log 2(iii)3 log 4
(iv)2 log 3 - 3 log 2(v)log243 + log1(vi)log 10 + 2 log 3 - log 2
5.Expand the following.
(i)log1000 (ii)log
128
625
 

 
(iii)logx
2
y
3
z
4
(iv)log
2 3
p q
r
(v)log
3
2
x
y
1.5.4 STANDARD BASES OF A LOGARITHM (NOT MEANT FOR EXAMINATION PURPOSE)
There are two bases which are used more commonly than any others and deserve
special mention. They are base 10 and base e
Usually the expression log x implies that the base is 10. In calculators, the button
marked log is pre-programmed to evaluate logarithms to base ‘10’.
For example,
log 2 = 0.301029995664…
log 3 = 0.4771212547197…
Is log 2 and log 3 irrational?

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22
The second common base is ‘e’. The symbol ‘e’ is called the exponential constant. This
is an irrational number with an infinite, non-terminating non-recurring decimal expansion. It is
usually approximated as 2.718. Base ‘e’ is used frequently in scientific and mathematical
applications. Logarithms to base e or log
e
, are often written simply as ‘ln’. So, “ln x” implies
the base is ‘e’. Such logarithms are also called natural logarithms. In calculators, the button
marked ‘ln’ gives natural logs.
For example
ln(2) = 0.6931471805599…
ln(3) = 1.0986122886681…
1.5.5 APPLICATION OF LOGARITHM S
Let us understand applications of logarithms with some examples.
Example-15. The magnitude of an earthquake was defined in 1935 by Charles Richer with the
expression M=log
I
S
; where I is the intensity of the earthquake tremor and S is the intensity of
a “threshold earthquake”.
(a)If the intensity of an earthquake is 10 times the intensity of a threshold earthquake, then
what is its magnitude?
(b)If the magnitude of an earthquake registers 10 on the Richter scale, how many times is
the intensity of this earthquake higher than that of a threshold earthquake?
Solution :
(a)Let the intensity of the earthquake be I, then we are given
I = 10 S
The magnitude of an earthquake is given by-
M = log
I
S
The magnitude of the Delhi earthquake will be-
M = log
I
S
= log 10
= 1
Is ln(2) and ln(3) irrational?

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23
(b)Let x be the number of times the intensity of the earthquake is higher than that of a
regualar earthquake. So the intensity of earthquake is-
I = xS
We know that-
M = log
I
S
So, the magnitude of the San Fransico earthquake is-
M = log
xs
s
or M = log x
We know thatM = 10
So log x = 10 and therefore x = 10
10
TRY THIS
The formula for calculating pH is pH = -log
10
[H+] where pH is the acidity of the solution
and H
+
is the hydrogen ion concentration.
(i)If Grandma’s Lux Soap has a hydrogen ion concentration of 9.2 × 10
(-12)
. What
is its pH?
(ii)If the pH of a tomato is 4.2, what is its hydrogen ion concentration?
OPTIONAL EXERCISE
[This exercise is not meant for examination]
1. Can the number 6
n
, n being a natural number, end with the digit 5? Give reason.
2. Is 7  5  3  2 + 3 a composite number? Justify your answer..
3. Check whether 12
n
can end with the digit 0 for any natural number n?
4. Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any
positive integer.
5. Prove that  2 3 5 is an irrational number. Also check whether
  2 3 5 2 3 5  is rational or irrational.

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24
6. Without actual division, find after how many places of decimals the decimal expansion of
the following rational numbers terminates. Verify by actual division. What do you infer?
(i)
5
16
(ii)
2
13
2
(iii)
17
125
(iv)
13
80
(v)
15
32
(vi)
2
33
2 5
7. If x
2
+ y
2
= 6xy, prove that 2 log (x + y) = logx + logy + 3 log 2
8. Find the number of digits in 4
2013
, if log
10
2 = 0.3010
WHAT WE HAVE DISCUSSED
1. The Fundamental Theorem of Arithmetic states that every composite number can be
expressed (factorized) as a product of its primes, and this factorization is unique, apart
from the order in which the prime factors occur.
2. If p is a prime and p divides a
2
, where a is a positive integer, then p divides q.
3. Let x be a rational number whose decimal expansion terminates. Then we can express x
in the form
p
q
, where p and q are coprime, and the prime factorization of q is of the form
2
n
5
m
, where n, m are non-negative integers.
4. Let x =
p
q
be a rational number, such that the prime factorization of q is of the form
2
n
5
m
, where n, m are non-negative integers. Then x has a decimal expansion which
terminates.
5. Let x =
p
q
be a rational number, such that the prime factorization of q is of the form
2
n
5
m
, where n, m are non-negative integers. Then x has a decimal expansion which is
non-terminating, repeating (recurring).
6. Logarithms are used for all sorts of calculations in engineering, science, business and
economics.
7. We define log
a
x = n, if a
n
= x, where a and x are positive numbers and a  1.
8. Laws of logarithms :
(i)log
a
xy = log
a
x + log
a
y (ii)log
a

x
y
= log
a
x  log
a
y
(iii)log
a
x
m
= m log
a
x

26. 2.1 INTRODUCTION
Look at the examples given below:
1. Euclid, Pythagoras, Gauss, Leibnitz, Aryabhata, Bhaskar.
2. a,e,i,o,u
3. Happy, sad, angry, anxious, joyful, confused.
4. Cricket, football, kho-kho, kabaddi, basketball.
5. 1, 3, 5, 7, 9.....
What do you observe? Example 1 is a collection of names of some mathematicians,
example 3 is a collection of feelings and example 2 is the collection of vowel letters in the English
alphabet. We see that the names/ items/ objects in each example have something in common., i.e.
they form a collection. Can you tell what are the collections in examples 4 and 5?
We come across collections in mathematics too. For example, natural numbers, prime
numbers, quadrilaterals in a plane etc. All examples seen so far are well defined collection of
objects or ideas. A well defined collection of objects or ideas is known as a set. Set theory is a
comparitively new concept in mathematics. It was developed by Georg Cantor (1845-1918).
In this chapter, we will learn about sets and their properties, and what we mean when we say
well-defined, elements of a set, types of sets etc.
2.2 WELL DEFINED SETS
What do we mean when we say that a set is a well defined collection of objects. Well
defined means that:
1. All the objects in the set should have a common feature or property; and
2. It should be possible to decide whether any given object belongs to the set or not.
Let us understand 'well defined' through some examples. Consider the statement : The
collection of all tall students in your class.
Sets2

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Mathematic26
What difficulty is caused by this statement? Here, who is tall is not clear. Richa decides that
all students taller than her are tall. Her set has five students. Yashodhara also decides that tall
means all students taller than her. Her set has ten students. Ganpati decides that tall means every
student whose height is more than 5 feet. His set has 3 students. We find that different people get
different collections. So, this collection is not well defined.
Now consider the following statement : The collection of all students in your class who are
taller than 5 feet 6 inches.
In this case, Richa, Yashodhara and Ganpati, all will get the same set. So, this collection
forms a well defined set.
DO THESE
1.Write 3 examples of 'sets' from your daily life.
2.Some collections are given below. Tick the ones that form well defined sets.
(i) Collection of all good students in your class.
(ii)Red, blue, green, yellow, purple.
(iii)1,2,3,4,5,6,7,.... (iv)1, 8, 27, 64, 125, ....
TRY THIS
State which of the following collections are sets.
(i)All even numbers (ii)Stars in the sky
(iii)The collection of odd positive integers. 1, 3, 5, …..
2.3 NAMING OF SETS AND ELEMENTS OF A SET
We usually denote a set by upper case letters, A, B, C, X, Y, Z etc. A few examples of sets
in mathematics are given below.
-The set of all Natural numbers is denoted by N.
-The set of all Integers is denoted by Z.
-The set of all Rational numbers is denoted by Q.
-The set of all Real numbers is denoted by R.
Notice that all the sets given above are well defined collections because given a number we
can decide whether it belongs to the set or not. Let us see some more examples of elements.

28. Free Distribution by A.P. Government
Sets 27
Suppose we define a set as all days in a week, whose name begins with T. Then we know
that Tuesday and Thursday are part of the set but Monday is not. We say that Tuesday and
Thursday are elements of the set of all days in a week starting with T.
Consider some more examples:
(i)We know that N usually stands for the set of all natural numbers. Then 1,2, 3... are
elements of the set. But 0 is not an element of N.
(ii)Let us consider the set, B of quadrilaterals
B = {square, rectangle, rhombus, parallelogram}
Can we put triangle, trapezium or cone in the above set, B? No, a triangle and cone are
can not be members of B. But a trapezium can be a member of the set B.
So, we can say that an object belonging to a set is known as a member/ element / individual
of the set. We use the symbol to denote 'belongs to'. So 1  N means that 1 belongs to N.
Similarly 0 N means that 0 does not belong N.
There are various ways in which we can write sets. For example, we have the set of all
vowel letters in the English alphabet. Then, we can write:
(i)V = {a, e, i, o, u}. Here, we list down all the elements of the set between chain/ curly
brackets. This is called the roster form of writing sets. In roster form, all elements of the
set are written, separated by commas, within curly brackets.
(ii)V = {x : x is a vowel letter in the English alphabet}
or V = {x | x is a vowel letter in the English alphabet}
This way of writing a set is known as the set builder form. Here , we use a symbol x ( or
any other symbol y, z etc.,) for the element of the set. This is is followed by a colon (or
a vertical line), after which we write the characteristic property possessed by the elements
of the set. The whole is enclosed within curly brackets.
LetC = {2, 3, 5, 7, 11}, a set of prime numbers less than 13. This set can be denoted as:
C = {x | x is a prime number less than 13}or
C = {x: x is a prime number less than 13}.
Example-1. Write the following in roster and set builder forms.
(i)The set of all natural numbers which divide 42.
(ii)The set of natural numbers which are less than 10.

29. Free Distribution by A.P. Government
Mathematic28
Solution :
(i)Let B be the set of all natural numbers which divide 42. Then, we can write:
B = {1, 2, 3, 6, 7, 14, 21, 42} Roster form
B = {x : x is a natural number which divides 42}Set builder form
(ii)Let A be the set of all natural numbers which are less than 10. Then, we can write:
A = {1, 2, 3, 4, 5, 6, 7, 8, 9}Roster form
B = {x : x is a natural number which is less than 10}Set builder form
Note :(i)In roster form, the order in which the elements are listed is immaterial. Thus, in
example 1, we can also write {1, 3, 7, 21, 2, 6, 4, 42}.
(ii)While writing the elements of a set in roster form, an element is not repeated. For
example, the set of letters forming the word “SCHOOL” is {S, C, H, O, L} and
not {S, C, H, O, O, L}
Example-2. Write the set B = {x : x is a natural number and x
2
< 40} in the roster form.
Solution : We look at natural numbers and their squares starting from 1. When we reach 7, the
sqaure is 49 which is greater than 40. The required numbers are 1, 2, 3, 4, 5, 6.
So, the given set in the roster form is B = {1, 2, 3, 4, 5, 6}.
DOTH ES E
1. List the elements of the following sets.
(i)G = all the factors of 20
(ii)F = the multiples of 4 between 17 and 61 which are divisible by 7
(iii)S = {x : x is a letter in the word 'MADAM'}
(iv)P = {x : x is a whole number between 3.5 and 6.7}
2. Write the following sets in the roster form.
(i)B, the set of all months in a year having 30 days.
(ii)P is the set of all prime numbers less than 10.
(iii)X = The colours of the rainbow
3. A is the set of factors of 12. Which one of the following is not a member of A.
(A) 1 (B) 4 (C) 5 (D) 12

30. Free Distribution by A.P. Government
Sets 29
TRY THESE
1. Formulate sets of your choice, involving algebraic and geometrical ideas.
2. Match roster forms with the set builder form.
(i){P, R, I, N, C, A, L}(a){x : x is a positive integer and is a divisor of 18}
(ii){0} (b){x : x is an integer and x
2
– 9 = 0}
(iii){1, 2, 3, 6, 9, 18}(c){x : x is an integer and x + 1 = 1}
(iv){3, -3} (d){x : x is a letter of the word PRINCIPAL}
EXERCISE - 2.1
1. Which of the following are sets? Justify your answer?
(i)The collection of all the months of a year beginning with the letter “J”.
(ii)The collection of ten most talented writers of India.
(iii)A team of eleven best cricket batsmen of the world.
(iv)The collection of all boys in your class.
(v)The collection of all even integers.
2. If A={0, 2, 4, 6}, B = {3, 5, 7} and C = {p, q, r}then fill the appropriate symbol,  or
 in the blanks.
(i)0 ….. A (ii) 3 ….. C (iii) 4 ….. B
(iv) 8 ….. A (v) p ….. C (vi) 7 ….. B
3. Express the following statements using symbols.
(i)The elements ‘x’ does not belong to ‘A’.
(ii)‘d’ is an element of the set ‘B’.
(iii)‘1’ belongs to the set of Natural numbers N.
(iv)‘8’ does not belong to the set of prime numbers P.
4. State whether the following statements are true or not.
(i)5 {Prime numbers}
(ii)S = {5, 6, 7} implies 8  S.
(iii)-5  W where‘W’ is the set of whole numbers
(iv)
8
11
 Z where ‘Z’ is the set of integers.

31. Free Distribution by A.P. Government
Mathematic30
5. Write the following sets in roster form.
(i)B = {x : x is a natural number less than 6}
(ii)C = {x : x is a two-digit natural number such that the sum of its digits is 8}.
(iii)D = {x : x is a prime number which is a divisor of 60}.
(iv)E = {the set of all letters in the word BETTER}.
6. Write the following sets in the set-builder form.
(i){3, 6, 9, 12} (ii){2, 4, 8, 16, 32}
(iii){5, 25, 125, 625} (iv)(1, 4, 9, 25, ….. 100}
7. List all the elements of the following sets.
(i)A = {x : x is a natural number greater than 50 but less than 100}
(ii)B = {x : x is an integer, x
2
= 4}
(iii)D = {x : x is a letter in the word “LOYAL”}
8. Match the roster form with set-builder form.
(i){1, 2, 3, 6} (a){x : x is prime number and a divisor of 6}
(ii){2, 3} (b){x : x is an odd natural number less than 10}
(iii){M, A, T, H, E, I, C, S} (c){x : x is a natural number and divisor of 6}
(iv){1, 3, 5, 7, 9} (d){x : x is a letter of the word MATHEMATICS}
2.4 TYPES OF SET
Let us consider the following examples of sets:
(i)A = {x : x is natural number smaller than 1}
(ii)D = {x : x is a prime number divisible by 2}
How many elements are there in A and D? We find that there is no natural number which
is smaller than 1. So set A contains no elements or we say that A is an empty set.
Similarly, there are no prime numbers that are divisible by 2. So, D is also an empty set.
A set which does not contain any element is called an empty set, or a Null set, or a void
set. Empty set is denoted by the symbol  or { }.

32. Free Distribution by A.P. Government
Sets 31
Here are some more examples of empty sets.
(i)A = {x : 1 < x < 2, x is a natural number}
(ii)B = {x : x
2
– 2 = 0 and x is a rational number}
(iii)D = {x : x
2
= 4, x is odd}
Note :  and {0} are two different sets. {0} is a set containing the single element 0 while {} is
null set.
Now consider the following sets:
(i)A = {the students of your school}(ii)L = {p,q,r,s}
(iii)B = {x : x is an even number}(iv)J = {x : x is a multiple of 7}
Can you list the number of elements in each of the sets given above? In (i), the number of
elements will be the number of students in your school. In (ii), the number of elements in set L is
4. We find that it is possible to count the number of elements of sets A and L or that they contain
a finite number of elements. Such sets are called finite sets.
Now, consider the set B of all even numbers. We cannot count all of them i.e., we see that
the number in elements of this set is not finite. Similarly, all the elements of J cannot be listed. We
find that the number of elements in B and J is infinite. Such sets are called infinite sets.
We can draw any number on straight lines passing though a given point. So this set is
infinite. Similarly, it is not possible to find out the last even number or odd number among the
collection of all integers. Thus, we can say a set is infinite if it is not finite.
Consider some more examples :
(i)Let ‘W’ be the set of the days of the week. Then W is finite.
(ii)Let ‘S’ be the set of solutions of the equation x
2
– 16 = 0. Then S is finite.
(iii)Let ‘G’ be the set of points on a line. Then G is infinite.
Example-3. State which of the following sets are finite or infinite.
(i){x : x  N and (x  1) (x  2) = 0}(ii){x : x  N and x
2
= 4}
(iii){x : x  N and 2x  1 = 0} (iv){x : x  N and x is prime}
(v){x : x  N and x is odd}
Solution :
(i) x can take the values 1 or 2 in the given case. The set is {1,2}. Hence, it is finite.

33. Free Distribution by A.P. Government
Mathematic32
(ii)x
2
= 4, implies that x = +2 or 2. But x N or x is a natural number so the set
is{2}. Hence, it is finite.
(iii)Given set x = 
1
2
but x N so x = . Hence, it is finite.
(iv)The given set is the set of all prime numbers. There are infinitely many prime numbers.
Hence, set is infinite.
(v)Since there are infinite number of odd numbers, hence the set is infinite.
Now, consider the following finite sets :
A = {1, 2, 4}; B = {6, 7, 8, 9, 10}; C = {a, b, c, a}
Here,
Number of elements in set A = 3.
Number of elements in set B = 5.
Number of elements in set C = 3 (In the set C, the element ‘a’ is written twice. We know
that the elements of a given set should be district. So, the number of distinct elements in set C is
3).
The number of elements in a set is called the cardinal number of the set. The cardinal
number of the set A is denoted as n(A) = 3.
Similarly, n(B) = 5 and n(C) = 3.
Note : There are no elements in a null set. The cardinal number of that set is 0.  n() = 0
Example-4. If A = {1, 2, 3}; B = {a, b, c} then find n(A) and n(B).
Solution : The set A contains three distinct elements  n(A) = 3
and the set B contains three distinct elements  n(B) = 3
DO THESE
1. Which of the following are empty sets? Justify your answer.
(i)Set of integers which lie between 2 and 3.
(ii)Set of natural numbers that are less than 1.
(iii)Set of odd numbers that have remainder zero, when divided by 2.

34. Free Distribution by A.P. Government
Sets 33
2. State which of the following sets are finite and which are infinite. Give reasons for your
answer.
(i)A = {x : x  N and x < 100} (ii)B = {x : x N and x < 5}
(iii)C = {1
2
, 2
2
, 3
2
, …..} (iv)D = {1, 2, 3, 4}
(v){x : x is a day of the week}.
3. Tick the set which is infinite
(A) The set of whole number < 10(B) The set of prime number < 10
(C) The set of integer < 10 (D) The set of factors of 10
TRY THIS
1. Which of the following sets are empty sets? Justify your answer.
(i)A = {x : x
2
= 4 and 3x = 9}.
(ii)The set of all triangles in a plane having the sum of their three angles less
than 180.
2. B = {x : x + 5 = 5} is not an empty set. Why?
THINK AND DISCUSS
An empty set is a finite set. Is this statement true or false? Why?
EXERCISE - 2.2
1. State which of the following sets are empty.
(i)The set of straight lines passing through a point.
(ii)Set of odd natural numbers divisible by 2.
(iii){x : x is a natural number, x < 5 and x > 7}
(iv){x : x is a point common to any two parallel lines}
(v)Set of even prime numbers.
2. Which of the following sets are finite or infinite.
(i)The set of months in a year.(ii){1, 2, 3, …, 99, 100}
(iii)The set of prime numbers less than 99.

35. Free Distribution by A.P. Government
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3. State whether each of the following set is finite or infinite.
(i)The set of letters in the English alphabet.
(ii)The set of lines which are parallel to the X-Axis.
(iii)The set of numbers which are multiplies of 5.
(iv)The set of circles passing through the origin (0, 0).
2.5 USING DIAGRAMS TO REPRESENT SETS
If S is a set and x is an object then either x  S or x S. Every set can be represented
by a drawing a closed curve C where elements of C are
represented by points within C and elements not in the
set by points outside C. For example, the set
D = {1, 2, 3, 4} can be represented as shown below:
2.6 UNIVERSAL SET AND SUBSETS
Let us consider that a cricket team is to be selected from your school. What is the set
from which the team can be selected? It is the set of all students in your school. Now, we want to
select the hockey team. Again, the set from which the team will be selected is all the set of
students in your school. So, for selection of any school team, the students of your school are
considered as the universal set.
Let us see some more examples of universal sets:
(i)The set of real number is the universal set for number theory.
(ii)If we want to study the various groups of people of our state, universal set is the set of
all people in Andhra Pradesh.
(iii)If we want to study the various groups of people in our country, universal set is the set of
all people in India.
The universal set is denoted by ''. The
Universal set is usually represented by rectangles.
If the set of real numbers, R is the universal set
in number theory then what about rational and irrational
numbers?
Let us consider the set of rational numbers,
Q = {x : x =
p
q
, p, q  z and q 0}
1
3
2
 4 5
D
HOCKEY
TEAM
CRICKET
TEAM


36. Free Distribution by A.P. Government
Sets 35
which is read as ‘Q’ is the set of all numbers such that x equals
p
q
, where p and q are integers
and q is not zero. Then we know that every element of Q is also an element of R. So, we can say
that Q is a subset of R. If Q is a subset of R, then we write is as Q  R.
Note :It is often convenient to use the symbol ‘’ which means implies.
Using this symbol, we can write the definition of subset as follows:
A  B if a A  a  B, where A, B are two sets.
We read the above statement as “A is a subset of B if 'a' is an element of A implies that
'a' is also an element of B”.
Real numbers, R has many subsets. For example,
The set of natural numbers N = {1, 2, 3, 4, 5, ……}
The set of whole number W = {0, 1, 2, 3, ......}
The set of integers Z = {…., -3, -2, -1, 0, 1, 2, 3, …..}
The set of irrational numbers T, is composed of all real numbers that are not rational.
Thus, T = {x : x  R and x  Q} i.e., all real numbers that are not rational. e.g. 2, 5
and π.
Similarly,the set of natural numbers, N is a subset of the set of whole numbers W and
we can write N  W. Also W is a subet of R.
That isN  W and W  R
N  W  R
Some of the obvious relations among these subsets are N  Z  Q, Q  R, T  R, and N
 T.
Example-5. Consider a set of vowels letters , V = {a, e, i, o, u}. Also consider the set A, of
all letters in the English alphabet. A = {a, b, c, d, ….., z}. Identify the universal set and the
subset in the given example.
Solution : We can see that every element of set V is also an element A. But every element of A
is not a part of V. In this case, V is the subset of A.
In other words V  A since whenever a  V, then a  A.
N
R
N
R
W
N
W

37. Free Distribution by A.P. Government
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Note : Since the empty set  has no elements, we consider that  is a subset of every set.
If A is not a subset of B (A  B), that means there is at least one element in A that is not a
member of B.
Let us consider some more examples of subsets.
 The set C = {1, 3, 5} is a subset of D = {5, 4, 3, 2, 1}, since each number 1, 3, and 5
belonging to C also belongs to D.
 Let A = {a, e, i, o, u} and B = {a, b, c, d} then A is not a subset of B. Also B is not a
subset of A.
2.6.1 EQ UA L SETS
Consider the following sets.
A = {Sachin, Dravid, Kohli}
B = {Dravid, Sachin, Dhoni}
C = {Kohli, Dravid, Sachin}
What do you observe in the a three sets A, B and C? All the players that are in A are in
C but not in B. Thus, A and C have same elements but some elements of A and B are different.
So, the sets A and C are equal sets but sets A and B are not equal.
Two sets A and C are said to be equal if every element in A belongs to C and every
element in C belongs to A. If A and C are equal sets, than we write A = C.
Example-6. Consider the following sets:
A = {p, q, r} B = {q, p, r}
In the above sets, every element of A is also an element of B.  A  B.
Similarly every element of B is also in A. B  A.
Thus, we can also write that if B  A and A  B A = B. Here  is the symbol for
two way implication and is usually read as, if and only if (briefly written as “iff”).
Examples-7. If A = {1, 2, 3, ….} and N is a set of natural numbers, check whether A and N
are equal?
Solution : The elements are same in both the sets. Therefore, both A and N are the set of
Natural numbers. Therefore the sets A and N are equal sets or A = N.

38. Free Distribution by A.P. Government
Sets 37
Example-8. Consider the sets A = {p, q, r, s} and B = {1, 2, 3, 4}. Are they equal?
Solution : A and B do not contain the same elements. So, A B.
Example-9. Let A be the set of prime numbers less than 6 and P the set of prime factors of 30.
Check if A and P are equal.
Solution : The set of prime numbers less than 6, A = { 2,3,5}
The prime factors of 30 are 2, 3 and 5. So, P = { 2,3,5}
Since the elements of A are the same as the elements of P, therefore, A and P are equal.
Note : A set does not change if one or more elements of the set are repeated. For example, the
sets A = {1, 2,3} and B = {2, 2, 1, 3, 3} are equal, since each element of A is in B and vice versa.
Example-10. Show that the sets A and B are equal, where
A = {x : x is a letter in the word ‘ASSASSINATION’}
B = {x : x is a letter in the word STATION}
Solution : Given,A = {x : x is a letter in the word ‘ASSASSINATION’}
This set A can also be written as A = {A,S,I,N,T,O}since generally elements in a set are
not repeated.
Also given thatB = {x : x is a letter in the word STATION}
‘B’ can also be written as B = {A,S,I,N,T,O}
So, the elements of A nd B are same and A = B
EXERCISE - 2.3
1. Which of the following sets are equal?
(i)A = {x : x is a letter in the word FOLLOW}
(ii)B = {x : x is a letter in the word FLOW}
(iii)C = {x : x is a letter in the word WOLF}
2. Consider the following sets and fill up the blank in the statement given below with = or
so as to make the statement true.
A = {1, 2, 3}; B = {The first three natural numbers}
C = {a, b, c, d}; D = {d, c, a, b}
E = {a, e, i, o, u}; F = {set of vowels in English Alphabet}

39. Free Distribution by A.P. Government
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(i)A _ B (ii)A _ E (iii)C _ D
(iv)D _ F (v)F _ A (vi)D _ E
(vii)F _ B
3. In each of the following, state whether A = B or not.
(i)A = {a, b, c, d} B = {d, c, a, b}
(ii)A = {4, 8, 12, 16} B = {8, 4, 16, 18}
(iii)A = {2, 4, 6, 8, 10} B = {x : x is a positive even integer and x < 10}
(iv)A = {x : x is a multiple of 10}B = {10, 15, 20, 25, 30, …….}
Consider the set E = {2, 4, 6} and F = {6, 2, 4}. Note that E = F. Now, since each
element of E also belongs to F, therefore E is a subset of F. But each element of F is also an
element of E. So F is a subset of E. In this manner it can be shown that every set is a subset of
itself.
If A and B contain the same elements, they are equal i.e. A = B. By this
observation we can say that “Every set is subset of itself”.
DO THIS
1. A = {1, 2, 3, 4},B = {2, 4},C = {1, 2, 3, 4, 7},f = { }.
Fill in the blanks with  or  .
(i)A ….. B (ii)C ….. A (iii)B ….. A
(iv)A ….. C (v)B ….. C (vi) ….. B
2. State which of the following statement are true.
(i) { } =  (ii) = 0 (iii)0 = { 0 }
TRY THESE
1. A = {quadrilaterals}, B = {square, rectangle, trapezium, rhombus}. State whether
A  B or B A. Justify your answer.
2. If A = {a, b, c, d}. How many subsets does the set A have? (Remember null set and
equal sets)
(A) 5 (B) 6 (C) 16 (D) 65

40. Free Distribution by A.P. Government
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3. P is the set of factors of 5, Q is the set of factors of 25 and R is the set of factors of 125.
Which one of the following is false?
(A) P  Q(B) Q  R(C) R  P(D) P  R
4. A is the set of prime numbers less than 10, B is the set of odd numbers < 10 and C is the
set of even numbers < 10. How many of the following statements are true?
(i) A  B (ii) B  A (iii) A  C
(iv) C  A (v) B  C (vi) X  A
Consider the following sets:
A = {1, 2, 3}, B = {1, 2, 3, 4}, C = {1, 2, 3, 4, 5}
All the elements of A are in B  A  B.
All the elements of B are in C  B  C.
All the elements of A are in C  A  C.
That is, A  B, B  C  A  C.
Example-11. Consider the sets , A = {1, 3}, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the
symbol  or  between each of the following pair of sets.
(i)  ….. B(ii) A ….. B(iii) A ….. C(iv) B ….. C
Solution :(i)  B, as  is a subset of every set.
(ii)A  B, for 3  A but 3  B.
(iii)A  C as 1, 3 A also belong to C.
(iv)B  C as each element of B is also an element of C.
EXERCISE - 2.4
1. State which of the following statements are true given that. A = {1, 2, 3, 4}
(i) 2  A (ii) 2  {1, 2, 3, 4}
(iii) A  {1, 2, 3, 4} (iv) {2, 3, 4}  {1, 2, 3, 4}
2. State the reasons for the following :
(i){1, 2, 3, …., 10} {x : x  N and 1 < x < 10}
(ii){2, 4, 6, 8, 10} {x : x = 2n+1 and x  N}

41. Free Distribution by A.P. Government
Mathematic40
(iii){5, 15, 30, 45} {x : x is a multiple of 15}
(iv){2, 3, 5, 7, 9} {x : x is a prime number}
3. List all the subsets of the following sets.
(i) B = {p, q} (ii) {x, y} (iii) D = {a, b, c, d}
(iv) [6, 12] (v) (6, 12]
2.7 VENN DIAGRAMS
We have already see some ways of representing sets using diagrams. Let us study it in
more detail now. Venn-Euler diagram or simply Venn-diagram is a way of representing the
relationships between sets. These diagrams consist of rectangles and closed curves usually
circles.
As mentioned earlier in the chapter, the universal set is usually represented by a rectangle.
(i)Consider that ={1, 2, 3, …., 10}
is the universal set of which,
A = {2, 4, 6, 8, 10} is a subset.
Then the venn-diagrams is as:
(ii) = {1, 2, 3, …., 10} is the universal
set of which, A = {2, 4, 6, 8, 10} and
B = {4, 6} are subsets and also B 
A. Then we have the following figure:
(iii)Let A = {a, b, c, d} and B = {c, d, e, f}.
Then we illustrate these sets with a Venn diagram as
2.8 BASIC OPERATIONS ON SETS
We know that arithmetic has operations of additions, subtraction and multiplication of
numbers. Similarly in sets, we define the operation of union, intersection and difference of sets.
1 A 3
2
4
6
10
5
9
7
8

1
9
3
5
7
A
B
2
8
10
 4
 6
a
b
e
f

c
d

42. Free Distribution by A.P. Government
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2.8.1 UNI ON OF SETS
Example-12. Suppose A is the set of students in your class who were absent on Tuesday and
B the set of students who were absent on Wednesday. Then,
A = {Sita, Hanif, Jaya} and
B = {Hanif, Ikram, Preeti}
Now, we want to find K, the set of students who were absent on either Tuesday or
Wednesday. Then, does Sita Hanif  Jaya  Ikram  Preeti  Ganpati 

Sita, Hanif, Jaya, Ikram and Preeti all belong to K but Ganpati does not.
Hence, K = {Sita, Hanif, Jaya, Ikram, Preeti}
Here K is the called the union of sets A and B. The union of A and B is the set which
consists of all the elements of A and B, the common elements being taken only once. The symbol
‘’ is used to denote the union. Symbolically, we write A  B and usually read as ‘A union B’.
A  B = {x : x  A or x  B}
Example-13. Let A = {2, 5, 6, 8} and B = {5, 7, 9, 1}. Find A  B.
Solution : We have A  B = {1, 2, 5, 6, 7, 8, 9}.
Note that the common element 5 was taken only once while writing A  B.
Example-14. Let A = {a, e, i, o, u} and B = {a, i, u}. Show that A  B = A.
Solution : We have A  B = {a, e, i, o, u} = A.
This example illustrates that union of sets A and its subset B is the Set A itself.
i.e, if B  A, then A  B = A.
The union of the sets can be represented by a Venn-diagram as shown (shaded portion)
Example-15. Illustrate A  B in Venn-diagrams where.
A = {1, 2, 3, 4} and B = {2, 4, 6, 8}
A B


43. Free Distribution by A.P. Government
Mathematic42
Solution :
2 . 8 . 2 INTERSECTION OF SETS
Let us again consider the example of absent students. This time we want to find the set L of
students who were absent on both Tuesday and Wednesday. We find that L = {Hanif}. Here, L
is called the intersection of sets A and B.
In general, the intersection of sets A and B is the set of all elements which are common to
A and B. i.e., those elements which belong to A and also belong to B. We denote intersection by
A  B. (read as “A intersection B”). Symbolically, we write
A  B = {x : x  A and x  B}
The intersection of A and B can be illustrated in the
Venn-diagram as shown in the shaded portion in the
adjacent figure.
Example-16. Find A  B when A = {5, 6, 7, 8} and B = {7, 8, 9, 10}.
Solution : The common elements in both A and B are 7, 8.
 A  B = {7, 8}.
Example-17. Illustrate A  B in Venn-diagrams where A = {1, 2, 3} and B = {3, 4, 5}
Solution : The intersection of A and B can be illustrated in the Venn-diagram as follows:
A B
1
3
2
2
6
8
A  B = {1, 2, 3, 4, 6, 8}

A B
A  B = {3}3
1
2
4
5


A  B
A B

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Sets 43
2 . 8 . 3 DISJOINT SET
Suppose A {1, 3, 5, 7} and B = {2, 4, 6, 8}. We
see that there are no common elements in A and B. Such
sets are known as disjoint sets. The disjoint sets can be
represented by means of the Venn-diagram as follows:
DO THESE
1. Let A = {1, 3, 7, 8} and B = {2, 4, 7, 9}. Find A  B.
2. If A = {6, 9, 11}; B = { }, find A  .
3. A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; B = {2, 3, 5, 7}. Find A  B and show that
A  B = B.
4. If A = {4, 5, 6}; B = {7, 8} then show that A  B = B  A.
TRY THESE
1. List out some sets A and B and choose their elements such that A and B are disjoint
2. If A = {2, 3, 5}, find A   and   A and compare.
3. If A = {1, 2, 3, 4}; B = {1, 2, 3, 4, 5, 6, 7, 8} then find A  B, A  B. What do you
notice about the result?
4. A = {1, 2, 3, 4, 5, 6}; B = {2, 4, 6, 8, 10}. Find the intersection of A and B.
THINK AND DISCUSS
The intersection of any two disjoint sets is a null set. Is this statement true or false?
2 . 8 . 4 DIFFERENCE OF SETS
The difference of sets A and B is the set of elements which belong to A but do not belong
to B. We denote the difference of A and B by A – B or simply “A Minus B”.
A – B = {x : x  A and x  B}.
Example-18. Let A = {1, 2, 3, 4, 5}; B = {4, 5, 6, 7}. Find A – B.
A
B


45. Free Distribution by A.P. Government
Mathematic44
Solution : Given A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}. Only the elements which are in A but
not in B should be taken.
 A – B = {1, 2, 3}. Since 4, 5 are elements in B they are not taken.
Similarly for B – A, the elements which are only in B are taken.
 B – A = {6, 7} (4, 5 are elements in A).
Note thatA – B  B – AA
The Venn diagram of A-B is as shown.
Example-19. Observe the following
A = {3, 4, 5, 6, 7}  n(A) = 5
B = {1, 6, 7, 8, 9}  n(B) = 5
A  B = {1, 3, 4, 5, 6, 7, 8, 9}n(A  B) = 8
A  B = {6, 7}  n(A  B) = 2
 n(A  B) = 5 + 5 – 2 = 8
We observe that n(A  B) = n(A) + n(B) – n(A  B)
DO THESE
1. If A = {1, 2, 3, 4 ,5}; B = {4, 5, 6, 7}then find A – B and B – A. Are they equal?
2. If V = {a, e, i, o, u} and B = {a, i,k, u}, find V – B and B – V.
3. If A = {9, 4, 3, 4, 3, 6}, B = {4, 5, 6, 6, 7, 9}, find n(A) and n(B).
THINK AND DISCUSS
The sets A – B, B – A and A  B are mutually disjoint sets. Sets are mutually disjoint if
the intersection of any of the two sets is the null set. Use examples to see if this is true.
A B
1
2
6
7
4
5
3
A B
1
2
4
5
3
A – B
6
7
B – A
3
4
5
6
7
1
8
9
A B 

46. Free Distribution by A.P. Government
Sets 45
EXERCISE - 2.5
1. If A = {1, 2, 3, 4}; B = {1, 2, 3, 5, 6} then find A  B and B  A. Are they same?
2. A = {0, 2, 4}, find A   and A  A. Comment.
3. If A = {2, 4, 6, 8, 10} and B = {3, 6, 9, 12, 15}, find A – B and B – A.
4. If A and B are two sets such that A  B then what is A  B?
5. If A = {x : x is a natural number}
B = {x : x is an even natural number}
C = {x : x is an natural number}
D = {x : x is a prime number}
Find A  B, A  C, A  D, B  C, B  D, C  D.
6. If A = {3, 6, 9, 12, 15, 18, 21};B = {4, 8, 12, 16, 20}
C = {2, 4, 6, 8, 10, 12, 14, 16}; D = {5, 10, 15, 20} find
(i) A – B(ii) A – C(iii) A – D(iv) B – A(v) C – A
(vi) D – A(vii) B – C(viii) B – D(ix) C – B(x) D – B
7. State whether each of the following statement is true or false. Justify you answers.
(i){2, 3, 4, 5} and {3, 6} are disjoint sets.
(ii){a, e, i, o, u} and {a, b, c, d} are disjoint sets.
(iii){2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
(iv){2, 6, 10} and {3, 7, 11} are disjoint sets.
WHAT WE HAVE DISCUSSED
1. A set is a well defined collection of objects where well defined means that:
(i)All the objects in the set have a common feature or property; and
(ii)It is possible to decide whether any given object belongs to the set or not.
2. An object belonging to a set is known as an element of the set. We use the symbol to
denote 'belongs to'.

47. Free Distribution by A.P. Government
Mathematic46
3. Sets can be written in the roster form where all elements of the set are written, separated
by commas, within curly brackets.
4. Sets can also be written in the set-builder form.
5. A set which does not contain any element is called an empty set, or a Null set, or a void
set.
6. A set is called a finite set if it is possible to count the number of elements of that set.
7. We can say that a set is infinite if it is not finite.
8. The number of elements in a set is called the cardinal number of the set.
9. The universal set is denoted by 'U'. The Universal set is usually represented by rectangles.
10.A is a subset of B if 'a' is an element of A implies that 'a' is also an element of B. This is
written as A  B if a A  a  B, where A, B are two sets.
11.Two sets, A and B are said to be equal if every element in A belongs to B and every
element in B belongs to A.
12.A union B is written as A  B = {x : x  A or x  B}.
13.A intersection B is written as A  B = {x : x  A and x  B}
14.Venn diagrams are a convenient way of showing operations between sets.

49. 3.1 INTRODUCTION
Let us observe two situations
1. A flower bed in a garden is in the shape of a triangle. The longest side is 3 times the
middle side and smallest side is 2 units shorter than the middle side. Let P represent the
length of the middle side, then what’s the perimeter in terms of P.
2. The length of a rectangular dining hall is twice its breadth. Let x represent the breadth of
the hall. What is the area of the floor of the hall in terms of x.
In the above situations, there is an unknown in each statement. In the first situation,
middle side is given as ‘P’ units.
Since, Perimeter of triangle = sum of all sides
Perimeter= P + 3P + P – 2
= 5P – 2
Similarly in the second situation, length is given as twice the breadth.
So, if breadth = x, length = 2x
Since area of rectangle = lb
Area = (2x) (x)
= 2 x
2
As you know, the perimeter, 5P – 2 of the triangle and area 2x
2
of the rectangle are in the form
of polynomials of different degrees.
3P
P-2
P
2x
2 x
2 x
Polynomials3

50. Mathematic48
Free Distribution by A.P. Government
3.2 WHAT ARE POLYNOMIALS?
Polynomials are algebraic expressions constructed using constants and variables.
Coefficients operate on variables, which can be raised to various powers of non-negative integer
exponents. For example, 2x + 5, 3x
2
+ 5x + 6, - 5y, x
3
are some polynomials.

3
2
1 1 1
, , , 3
12
x
x yx
etc. are not polynomials.
Why is
1
1y
not a polynomial? Discuss with your friends and teacher..
DO THIS
State which of the following are polynomials and which are not? Give reasons.
(i) 2x
3
(ii)
1
1x
(iii)
21
4
7
z (iv)
2
2 2m m  (v)
2
1P


3 . 2 . 1 DEGREE OF A POLYNOMIAL
Recall that if p(x) is a polynomial in x, the highest power of x in p(x) is called the degree
of the polynomial p(x). For example, 3x + 5 is a polynomial in the variable x. It is of degree 1 and
is called a linear polynomial. 5x, 2 5y ,
1
3
P, , m + 1 etc. are some more linear polynomials.
A polynomial of degree 2 is called a quadratic polynomial. For example, x
2
+ 5x + 4 is
a quadratic polynomial in the variable x. 2x
2
+ 3x -
1
2
, p
2
-1, 3 – z – z
2
, y
2
-
3
y
+ 2 are some
examples of quadratic polynomials.
The expression 5x
3
–4x
2
+x–1 is a polynomial in the variable x of degree 3, and is called
a cubic polynomial. Some more examples of cubic polynomials are 2 – x
3
, p
3
, 
3
– 
2
–  + 5.
TRY THIS
Write 3 different quadratic, cubic and 2 linear polynomials with different number of terms.
We can write polynomials of any degree. 7 u
6
–
3
2
u
4
+ 4 u
2
– 8 is polynomial of degree
6. x
10
– 3x
8
+ 4x
5
+ 2x
2
-1 is a polynomial of degree 10. We can write a polynomial in a
variable x of a degree n where n is any natural number.

51. Polynomials49
Free Distribution by A.P. Government
Generally, we say
p(x) = a
0
x
n
+ a
1
x
n-1
+ a
2
x
n-2
+ …….. + a
n-1
x +a
n
be a polynomial of n
th
degree,
where a
0
, a
1
, a
2
….. a
n-1,
a
n
are real coefficients and a
0


0
For example, the general form of a first degree polynomial in one variable x is ax+b,
where a and b are real numbers and a  0.
TRY THESE
1.Write a quadratic polynomial and a cubic polynomial in variable x in the general form.
2.Write a general polynomial q(z) of degree n with coefficients that are b
0
...b
n
. What are the
conditions on b
0
...b
n
?.
3.2.2 VALUE OF A POLYNOM IAL
Now consider the polynomial p(x) = x
2
– 2x – 3. What is the value of the polynomial at
any point? For example, what is the value at x = 1? Putting x = 1, in the polynomial, we get p(1)
= (1)
2
– 2(1) – 3 = –4. The value – 4, is obtained by replacing x by 1 in the given polynomial
p(x). This is the value of x
2
– 2x – 3 at x = 1.
Similarly, p(0) = –3 is the value of p(x) at x = 0.
Thus, if p(x) is a polynomial in x, and if k is a real number, then the value obtained by
replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k).
DO THIS
(i)p(x) = x
2
– 5x – 6, find the values of p(1), p(2), p(3), p(0), p(–1), p(–2), p(–3).
(ii)p(m) = m
2
– 3m + 1, find the value of p(1) and p(–1).
3 . 2 . 3 ZEROES OF A POLYNOMIAL
What is the value ofp(x) = x
2
– 2x -3 at x =3
We have, p(3) = (3)
2
– 2(3) – 3 = 9 – 6- 3 = 0
Also p(-1) = (-1)
2
– 2(-1) - 3 = 1 + 2 – 3 = 0
and p(2) = (2)
2
– 2(2) – 3 = 4 – 4 – 3 = -3

52. Mathematic50
Free Distribution by A.P. Government
We see that p(3) = 0 and p(-1) = 0. These points, x = 3 and x = –1, are called Zeroes
of the polynomial p(x) = x
2
– 2x -3.
As p(2) 

0, 2 is not the zero of p(x).
More generally, a real number k is said to be a zero of a polynomial p(x), if p(k) = 0.
DO THIS
(i)Let p(x) = x
2
– 4x + 3. Find the value of p(0), p(1), p(2), p(3) and obtain zeroes of the
polynomial p(x).
(ii)Check whether -3 and 3 are the zeroes of the polynomial x
2
– 9.
EXERCISE - 3.1
1. (a)If p(x) = 5x
7
– 6x
5
+ 7x-6, find
(i) coefficient of x
5
(ii) degree of p(x) (iii) constant term.
(b) Write three more polynomials and create three questions for each of them.
2. State which of the following statements are true and which are false? Give reasons for
your choice.
(i)The degree of the polynomial 2x
2
– 3x + 1 is 2.
(ii)The coefficient of x
2
in the polynomial p(x) = 3x
3
– 4x
2
+ 5x + 7 is 2.
(iii)The degree of a constant term is zero.
(iv) 2
1
5 6x x 
is a quadratic polynomial.
(v)The degree of a polynomial is one more than the number of terms in it.
3.If p(t) = t
3
– 1, find the values of p(1), p(–1), p(0), p(2), p(–2).
4.Check whether –2 and 2 are the zeroes of the polynomial x
4
– 16
5.Check whether 3 and –2 are the zeroes of the polynomial p(x) when p(x) = x
2
– x – 6.
3.3 WORKING WITH POLYNOMIALS
You have already studied how to find the zeroes of a linear polynomial.

53. Polynomials51
Free Distribution by A.P. Government
For example, if k is a zero of p(x) = 2x + 5, then p(k) =0 gives 2k +5 = 0 i.e., k =
5
2

.
In general, if k is a zero of p(x) = ax+b,
then p(k) = ak + b =0,
ie. k =
b
a

, or the zero of the linear polynomial ax + b is
b
a

.
Thus, the zero of a linear polynomial is related to its coefficients, including the constant
term.
Are the zeroes of higher degree polynomials also related to their coefficients? Think
about this and discuss with friends. We will come back to this later.
3.4 GEOMETRICAL MEANING OF THE ZEROES OF A POLYNOMIAL
You know that a real number k is a zero of the polynomial p(x) if p(k) = 0. Let us see the
graphical representations of linear and quadratic polynomials and the geometrical meaning of
their zeroes.
3.4.1. GRAPHICAL REPRESENTATION OF A LINEAR POLY NOMIAL
Consider first a linear polynomial ax + b, a  0. You have studied in Class-IX that the graph of
y = ax + b is a straight line. For example, the graph of y = 2x + 3 is a straight line intersecting the
y-axis at y = 3 and passing through the points (–2, –1) and (2, 7).
Table 3.1
x –2 0 2
y = 2x + 3 –1 3 7
(x, y) (–2, –1) (0, 3) (2, 7)

54. Mathematic52
Free Distribution by A.P. Government
From the graph, you can
see that the graph of y = 2x +
3 intersects the x-axis between
x = –1 and x = –2, that is, at
the point x =
3
2

. But x =
3
2

is also the zero of the
polynomial 2x + 3. Thus, the
zero of the polynomial
2x + 3 is the x-coordinate of
the point where the graph of
y = 2x + 3 intersects the
x-axis.
DO THIS
Draw the graph of (i) y = 2x + 5, (ii) y = 2x – 5, (iii) y = 2x and find the point of intersection
on x-axis. Is this also the zero of the polynomial?
In general, for a linear polynomial ax + b, a  0, the graph of y = ax + b is a straight line
which intersects the x-axis at exactly one point, namely, ,0
b
a
 

  
.
Therefore, the linear polynomial ax + b, a  0, has exactly one zero, namely, the x-
coordinate of the point where the graph of y = ax + b intersects the x-axis.
3.4.2. GRAPHICAL REPRESENTATION OF A QUADRATIC POLYNOMIAL
Now, let us look for the geometrical meaning of a zero of a quadratic polynomial. Consider
the quadratic polynomial x
2
– 3x – 4. Let us see what the graph of y = x
2
– 3x – 4 looks like. Let
us list a few values of y = x
2
– 3x – 4 corresponding to a few values for x as given in Table 3.2.
1
2
3
4
5
6
7
-1
-1
-2-2
-3-3
-4-4
-5
-6
-7
0
1234567-2-3-4-5-6
X' X
-7
Y
Y'
(-2, -1)
3
,0
2
 
 
 
(0, 3)
(2, 7)

55. Polynomials53
Free Distribution by A.P. Government
Table 3.2
x – 2 – 1 0 1 2 3 45
y = x
2
– 3x – 46 0 – 4– 6 – 6– 4 06
(x, y) (– 2, 6)(– 1, 0)(0, 4)(1, – 6)(2, – 6)(3, – 4)(4, 0)(5, 6)
We locate the points listed
above on a graph paper and
draw the graph.
Is the graph of this quadratic
polynomial a straight line? It is
like a shaped curve. It
intersects the x-axis at two
points.
In fact, for any quadratic
polynomial ax
2
+ bx + c,
a0, the graph of the
corresponding equation
y = ax
2
+ bx + c either opens
upwards like this or opens
downwards like . This
depends on whether a > 0 or
a < 0. (These curves are called parabolas.)
We observe that -1 and 4 are zeroes of the quadratic polynomial and -1 and 4 are
intersection points of x-axis. Zeroes of the quadratic polynomial x
2
– 3x – 4 are the x-coordinates
of the points where the graph of y = x
2
– 3x – 4 intersects the x-axis.
This is true for any quadratic polynomial, i.e., the zeroes of a quadratic polynomial ax
2
+
bx + c, a 0, are precisely the x-coordinates of the points where the parabola representing y =
ax
2
+ bx + c intersects the x-axis.
TRY THESE
Draw the graphs of (i) y = x
2
– x – 6 (ii) y = 6 – x – x
2
and find zeroes in each case.
What do you notice?
From our observation earlier about the shape of the graph of y = ax
2
+ bx + c, the
following three cases can happen:
1
2
3
4
5
6
7
-1
-1
-2-2
-3-3
-4-4
-5
-6
-7
0
1234567-2-3-4-5-6
X' X
-7
Y
Y'
(-2, 6)
(-1, 0)
(0, -4)
(1, -6)
(2, -6)
(3, -4)
(4, 0)
(5, 6)

56. Mathematic54
Free Distribution by A.P. Government
Case (i) : Here, the graph cuts x-axis at two distinct points A and A. In this case, the
x-coordinates of A and Aare the two zeroes of the quadratic polynomial ax
2
+bx +c . The
parabola can open either upward or downward.
(i) (ii)
Case (ii) : Here, the graph touches x-axis at exactly one point, i.e., at two coincident points. So,
the two points A and Aof Case (i) coincide here to become one point A.
(i) (ii)
The x-coordinate of A is the only zero for the quadratic polynomial ax
2
+ bx + c in this case.
X' X
Y
Y'
X' X
Y
Y'
A
A'O A' AO
X' X
Y
Y'
X' X
Y
Y'
A
O AO

57. Polynomials55
Free Distribution by A.P. Government
Case (iii) : Here, the graph is either completely above the x-axis or completely below the
x-axis. So, it does not cut the x-axis at any point.
(i) (ii)
So, the quadratic polynomial ax
2
+ bx + c has no zero in this case.
So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or
two equal zeroes (i.e., one zero), or no zero. This also means that a polynomial of degree 2 has
atmost two zeroes.
TRY THIS
1.Write three polynomials that have 2 zeros each.
2.Write one polynomial that has one zero.
3.How will you verify if it has only one zero.
4.Write three polynomials that have no zeroes for x that are real numbers.
3 . 4 . 3GEOMETRICAL MEANING OF ZEROES OF A CUBIC POLYNOMIAL
What could you expect the geometrical meaning of the zeroes of a cubic polynomial to
be? Let us find out. Consider the cubic polynomial x
3
– 4x. To see what the graph of y = x
3
– 4x
looks like, let us list a few values of y corresponding to a few values for x as shown in Table 3.6.
X' X
Y
Y'
O
X' X
Y
Y'
O

58. Mathematic56
Free Distribution by A.P. Government
Table 3.6
x –2 –1 0 1 2
y = x
3
– 4x 0 3 0 –3 0
(x, y) (–2, 0) (–1, 3) (0, 0) (1, –3)(2, 0)
On drawing the graph, we see
that the graph of y = x
3
– 4x
looks like the one given in the
figure.
We see from the table above
that –2, 0 and 2 are zeroes of
the cubic polynomial x
3
– 4x.
–2, 0 and 2 are x, the
x-coordinates of the points
where the graph of y = x
3
– 4x
intersects the x-axis. So this
polynomial has three zeros.
Let us take a few more
examples. Consider the cubic
polynomials x
3
and x
3
– x
2
respectively. See Table 3.7
and 3.8.
Table 3.7
x –2 –1 0 1 2
y = x
3
–8 –1 0 1 8
(x, y) (–2, –8)(–1, –1) (0, 0) (1, 1) (2, 8)
1
2
3
4
5
6
7
-1
-1
-2-2
-3-3
-4-4
-5
-6
-7
0
1234567-2-3-4-5-6
X' X
-7
Y
Y'
(1, -3)
(2, 0)
(-1, 3)
(-2, 0)

59. Polynomials57
Free Distribution by A.P. Government
Table 3.8
x –2 –1 0 1 2
y = x
3
– x
2
–12 –2 0 0 4
(x, y) (–2, –12)(–1, –2) (0, 0) (1, 0) (2, 4)
y = x
3
y = x
3
– x
2
In y = x
3
, you can see that 0 is the x-coordinate of the only point where the graph of y =
x
3
intersects the x-axis. So, the polynomial has only one distinct zero. Similarly, 0 and 1 are the
x-coordinates of the only points where the graph of y = x
3
– x
2
intersects the x-axis. So, the
cubic polynomial has two distinct zeros.
From the examples above, we see that there are at most 3 zeroes for any cubic polynomial.
In other words, any polynomial of degree 3 can have at most three zeroes.
TRY THESE
Find the zeroes of cubic polynomials (i) – x
3
(ii) x
2
– x
3
(iii) x
3
– 5x
2
+ 6x without drawing
the graph of the polynomial.
1
2
3
4
5
6
7
-1
-1
-2-2
-3-3
-4-4
-5
-6
-7
0
1234-2-3-4
X' X
Y
Y'
1
2
3
4
5
6
7
-1
-1
-2-2
-3-3
-4-4
-5
-6
-7
0
1234-2-3-4
X' X
Y
Y'
(-2, -8)
(-1, -1)
(2, 8)
(-1, -2)
(1, 0)
(2, 4)
(1, 1)
(0, 0)

60. Mathematic58
Free Distribution by A.P. Government
Remark : In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the
x-axis at at most n points. Therefore, a polynomial p(x) of degree n has at most n zeroes.
Example-1. Look at the graphs in the figures given below. Each is the graph of y = p(x), where
p(x) is a polynomial. In each of the graphs, find the number of zeroes of p(x) in the given range
of x.
Solution : In the given range of x in respective graphs :
(i)The number of zeroes is 1 as the graph intersects the x-axis at one point only.
(ii)The number of zeroes is 2 as the graph intersects the x-axis at two points.
(iii)The number of zeroes is 3. (Why?)
(iv)The number of zeroes is 1. (Why?)
(v)The number of zeroes is 1. (Why?)
(vi)The number of zeroes is 4. (Why?)
X' X
Y
Y'
X' X
Y
Y'
X' X
Y
Y'
O OO
X' X
Y
Y'
X' X
Y
Y'
X' X
Y
Y'
O OO
(i) (ii) (iii)
(iv) (v) (vi)

61. Polynomials59
Free Distribution by A.P. Government
Example-2. Find the number of zeroes of the given polynomials. And also find their values.
(i) p(x) = 2x + 1 (ii) q(y) = y
2
– 1 (iii) r(z) = z
3
Solution : We will do this without plotting the graph.
(i)p(x) = 2x + 1 is a linear polynomial. It has only one zero.
To find zeroes,
Let p(x) = 0
So, 2x+1=0
Therefore x =
1
2

The zero of the given polynomial is
1
2

.
(ii)q(y) = y
2
– 1 is a quadratic polynomial.
It has atmost two zeroes.
To find zeroes, let q(y) = 0
 y
2
– 1 = 0
 (y + 1) (y – 1) = 0
 y = -1 or y = 1
Therefore the zeroes of the polynomial are -1 and 1.
(iii)r(z) = z
3
is a cubic polynomial. It has at most three zeroes.
Let r(z) = 0
 z
3
= 0
So, the zero of the polynomial is 0.
EXERCISE – 3.2
1. The graphs of y = p(x) are given in the figure below, for some polynomials p(x). In each
case, find the number of zeroes of p(x).

62. Mathematic60
Free Distribution by A.P. Government
2. Find the zeroes of the given polynomials.
(i) p(x) = 3x (ii) p(x) = x
2
+ 5x + 6
(iii) p(x) = (x+2) (x+3) (iv) p(x) = x
4
– 16
3. Draw the graphs of the given polynomials and find the zeroes. Justify the answers.
(i) p(x) = x
2
– x – 12 (ii) p(x) = x
2
– 6x + 9
(iii) p(x) = x
2
– 4x + 5 (iv) p(x) = x
2
+ 3x – 4
(v) p(x) = x
2
– 1
4. Why are
1
4
and –1 zeroes of the polynomials p(x) = 4x
2
+ 3x – 1?
3.5 RELATIONSHIP BETWEEN ZEROES AND COEFFICIENTS OF A POLYNOMIAL
You have already seen that zero of a linear polynomial ax + b is –
b
a
. We will now try to
explore the relationship between zeroes and coefficients of a quadratic polynomial. For this, let
us take the quadratic polynomial p(x) = 2x
2
– 8x + 6.
X' X
Y
Y'
X' X
Y
Y'
X' X
Y
Y'
O OO
X' X
Y
Y'
X' X
Y
Y'
X' X
Y
Y'
O OO
(i) (ii) (iii)
(iv) (v) (vi)

63. Polynomials61
Free Distribution by A.P. Government
In Class-IX, we have learnt how to factorise quadratic polynomials by splitting the middle
term. So, here we split the middle term ‘–8x’ as a sum of two terms, whose product is 6 × 2x
2
=
12 x
2
. So, we write
2x
2
– 8x + 6= 2x
2
– 6x – 2x + 6 = 2x(x – 3) – 2(x – 3)
= (2x – 2) (x – 3) = 2(x – 1) (x – 3)
p(x) = 2x
2
– 8x + 6 is zero when x – 1 = 0 or x – 3 = 0, i.e., when x = 1 or x = 3. So,
the zeroes of 2x
2
– 8x + 6 are 1 and 3. We now try and see if these zeroes have some relationship
to the coefficients of terms in the polynomial. The coefficient of x
2
is 2; of x

is –8 and the
constant is 6, which is the coefficient of x
0
. (i.e. 6x
0
= 6)
We see that the sum of the zeroes = 1 + 3 = 4 =
( 8)
2
 
= 2
(coefficient of )
coefficientof
x
x

Product of the zeroes= 1 × 3 = 3 =
6
2
= 2
constant term
coefficient ofx
Let us take one more quadratic polynomial:
p(x) = 3x
2
+ 5x – 2.
By splitting the middle term we see,
3x
2
+ 5x – 2 = 3x
2
+ 6x – x – 2 = 3x(x + 2) – 1(x + 2)
= (3x – 1) (x + 2)
3x
2
+ 5x – 2 is zero when either 3x – 1 = 0 or x + 2 = 0
i.e., when x =
1
3
or x = –2.
The zeroes of 3x
2
+ 5x – 2 are
1
3
and –2. We can see that the :
Sum of its zeroes=
1
3
+ (-2) =
5
3

= 2
(coefficient of )
coefficientof
x
x

Product of its zeroes=
1
3
× (-2) =
2
3

= 2
constant term
coefficient ofx

64. Mathematic62
Free Distribution by A.P. Government
DO THIS
Find the zeroes of the quadratic polynomials given below. Find the sum and product of
the roots and verify relationship to the coefficients of terms in the polynomial.
(i) p(x) = x
2
– x – 6 (ii) p(x) = x
2
– 4x + 3
(iii) p(x) = x
2
– 4 (iv) p(x) = x
2
+ 2x + 1
In general, if  and  are the zeroes of the quadratic polynomial p(x) = ax
2
+ bx + c,
where a  0, then (x – ) and (x – ) are the factors of p(x). Therefore,
ax
2
+ bx + c = k (x – ) (x – ), where k is a constant
= k[x
2
– (x + ) x + ]
= k x
2
– k ( + ) x + k  
Comparing the coefficients of x
2
, x and constant terms on both the sides, we get
a = k, b = – k(+) and c = k.
This gives  +  =
b
a

,
 =
c
a
Note : and are Greek letters pronounced as ‘alpha’ and ‘beta’ respectively. We will
use later one more letter ‘’ pronounced as ‘gamma’.
So, sum of zeroes of a quadratic polynomial =  +  =
b
a

= 2
(coefficient of )
coefficientof
x
x

Product of zeroes of a quadratic polynomial =  =
c
a
= 2
constant term
coefficient ofx
Let us consider some examples.
Example-3. Find the zeroes of the quadratic polynomial x
2
+ 7x + 10, and verify the relationship
between the zeroes and the coefficients.
Solution : We have
x
2
+ 7x + 10 = (x + 2) (x + 5)

65. Polynomials63
Free Distribution by A.P. Government
So, the value of x
2
+ 7x + 10 is zero when x + 2 = 0 or x + 5 = 0,
i.e., when x = –2 or x = –5.
Therefore, the zeroes of x
2
+ 7x + 10 are –2 and –5.
Now, sum of the zeroes= –2 + (–5) = – (7) =
(7)
1

= 2
(coefficient of )
coefficientof
x
x

Product of the zeroes= –2 × (–5) = 10 =
10
1
= 2
constant term
coefficient ofx
Example-4. Find the zeroes of the polynomials x
2
– 3 and verify the relationship between the
zeroes and the coefficients.
Solution : Recall the identity a
2
– b
2
= (a – b) (a + b).
Using it, we can write:
x
2
– 3 = (x – 3) (x + 3)
So, the value of x
2
– 3 is zero when x = 3 or x = – 3.
Therefore, the zeroes of x
2
– 3 are 3 and –3 .
Sum of the zeroes= 3 – 3 = 0 = 2
(coefficient of )
coefficientof
x
x

Product of zeroes= (3) × (– 3) = – 3 =
3
1

= 2
constant term
coefficient ofx
Example-5. Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2,
respectively.
Solution : Let the quadratic polynomial be ax
2
+ bx + c, and its zeroes be  and . We have
 +  = – 3 =
b
a

,
and = 2 =
c
a
.
If we take a = 1, then b = 3 and c = 2
So, one quadratic polynomial which fits the given conditions is x
2
+ 3x + 2.

66. Mathematic64
Free Distribution by A.P. Government
Similarly, we can take 'a' to be any real number. Let us say it is k. This gives 3
b
k

 
or b = 3k and 2
c
k
 or c = 2k. Putting the values of a, b and c, we get the polynomial is kx
2
+
3kx + 2k.
Example-6. Find a quadratic polynomial if the zeroes of it are 2 and
1
3

respectively..
Solution : Let the quadratic polynomial be
ax
2
+ bx + c, a  0 and its zeroes be  and .
Here  = 2,  =
1
3

Sum of the zeroes = ( + ) = 2 +
1
3
 

 
=
5
3
Product of the zeroes= () = 2
1
3
 

 
=
2
3

Therefore the quadratic polynomial
ax
2
+ bx + c is k[x
2
– ( + )x +  ], where k is a constant
= k[x
2
–
5
3
x –
2
3
]
We can put different values of k.
When k = 3, the quadratic polynomial will be 3x
2
– 5x – 2.
TRY THESE
(i)Find a quadratic polynomial with zeroes 2 and
1
3
.
(ii)What is the quadratic polynomial whose sum of zeroes is
3
2

and the product
of zeroes is 1.

67. Polynomials65
Free Distribution by A.P. Government
3.6 CUBIC POLYNOMIALS
Let us now look at cubic polynomials. Do you think some relation holds between the
zeroes of a cubic polynomial and its coefficients as well?
Let us consider p(x) = 2x
3
– 5x
2
– 14x + 8.
We see that p(x) = 0 for x = 4, – 2,
1
2
.
Since p(x) can have at most three zeroes, these are the zeroes of 2x
3
– 5x
2
– 14x + 8.
Sum of its zeroes= 4 + (–2) +
1
2
=
5
2
=
( 5)
2
 
=
2
3
(coefficient of )
coefficient of
x
x

Product of its zeroes= 4 × (–2) ×
1
2
= – 4 =
8
2

= 3
(constant term)
coefficientofx

However, there is one more relationship here. Consider the sum of the products of the
zeroes taken two at a time. We have:
=
1 1
{4 ( 2)} ( 2) 4
2 2
         
         
   
      
= – 8 – 1 + 2 = – 7 =
14
2

= 3
constant of
coefficient of
x
x
In general, it can be proved that if , ,  are the zeroes of the cubic polynomial
ax
3
+ bx
2
+ cx + d, then
 +  +  =
b
a

,
 +  +  =
c
a
,
   =
d
a

.
3 2
ax bx cx d   is a polynomial with roots , ,  . Let us
see how , ,   relate to a, b, c.
Since , ,   are the zeroes, the polynomial can be written as
( )( )( )x x x    
3 2
( ) ( )x x x              
To compare with the polynomial, we multiply by 'a' and get
3 2
( ) ( )ax x a xa a             
or ( ), ( ),b a c a d a              

68. Mathematic66
Free Distribution by A.P. Government
DO THIS
If , ,  are the zeroes of the given cubic polynomials, find the values as given in the
table.
S.No. Cubic Polynomial +  +   +  +    
1 x
3
+ 3x
2
– x – 2
2 4x
3
+ 8x
2
– 6x – 2
3 x
3
+ 4x
2
– 5x – 2
4 x
3
+ 5x
2
+ 4
Let us consider an example.
Example-7. Verify that 3, –1, –
1
3
are the zeroes of the cubic polynomial
p(x) = 3x
3
– 5x
2
– 11x – 3, and then verify the relationship between the zeroes and the coefficients.
Solution : Comparing the given polynomial with ax
3
+ bx
2
+ cx + d, we get
a = 3, b = – 5, c = – 11, d = – 3 . Further
p(3) = 3 × 3
3
– (5 × 3
2
) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,
p(–1) = 3 × (–1)
3
– 5 × (–1)
2
– 11 × (–1) – 3 = – 3 – 5 + 11 – 3 = 0,
3 2
1 1 1 1
3 5 11 3
3 3 3 3
p
       
                                   
,
1 5 11 2 2
3 0
9 9 3 3 3
        
Therefore, 3, –1, and
1
3
 are the zeroes of 3x
3
– 5x
2
– 111x – 3.
So, we take  = 3,  = –1 and  =
1
3
.
Now,
 +  +  = 3 + (–1) +
1
3
 
 
= 2
1
3
 =
5
3
=
( 5)
3
 
=
b
a

,
 +  +  = 3 × (–1) + (–1) ×
1
3
 
 
+
1
3
 
 
× 3 = – 3 +
1
3
– 1 =
11
3

=
c
a
,
   = 3 × (–1) ×
1
3
 
 
= – 1 =
( 3)
3
 
=
d
a

.

69. Polynomials67
Free Distribution by A.P. Government
EXERCISE – 3.3
1. Find the zeroes of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.
(i) x
2
– 2x – 8 (ii) 4s
2
– 4s + 1 (iii) 6x
2
– 3 – 7x
(iv) 4u
2
+ 8u (v) t
2
– 15 (vi) 3x
2
– x – 4
2. Find the quadratic polynomial in each case, with the given numbers as the sum and
product of its zeroes respectively.
(i)
1
4
, – 1 (ii)
1
2,
3
(iii) 0, 5
(iv) 1, 1 (v) –
1
4
,
1
4
(vi) 4, 1
3. Find the quadratic polynomial, for the zeroes ,  given in each case.
(i) 2, –1(ii) 3, –3 (iii)
1
4
, – 1(iv)
1
2
,
3
2
4. Verify that 1, –1 and –3 are the zeroes of the cubic polynomial x
3
+ 3x
2
– x – 3 and
check the relationship between zeroes and the coefficients.
3.7 DIVISION ALGORITHM FOR POLYNOMIALS
You know that a cubic polynomial has at most three zeroes. However, if you are given
only one zero, can you find the other two? For this, let us consider the cubic polynomial
x
3
+ 3x
2
– x – 3. If we tell you that one of its zeroes is 1, then you know that this is divisible by
x – 1. Dividing by x – 1 we would get the quotient x
2
– 2x – 3.
We get the factors of x
2
– 2x – 3 by splitting the middle term. The factors are (x + 1) and
(x – 3). This gives us
x
3
+ 3x
2
– x – 3 = (x – 1) (x
2
– 2x – 3)
= (x – 1) (x + 1) (x – 3)
So, the three zeroes of the cubic polynomial are 1, – 1, 3.
Let us discuss the method of dividing one polynomial by another in some detail. Before
doing the steps formally, consider a particular example.

70. Mathematic68
Free Distribution by A.P. Government
Example-8. Divide 2x
2
+ 3x + 1 by x + 2.
Solution : Note that we stop the division process when either the
remainder is zero or its degree is less than the degree of the divisor. So,
here the quotient is 2x – 1 and the remainder is 3. Also,
(2x – 1) (x + 2) + 3 = 2x
2
+ 3x – 2 + 3 = 2x
2
+ 3x + 1
i.e., 2x
2
+ 3x + 1 = (x + 2) (2x – 1) + 3
Therefore, Dividend = Divisor × Quotient + Remainder
Let us now extend this process to divide a polynomial by a
quadratic polynomial.
Example-9. Divide 3x
3
+ x
2
+ 2x + 5 by 1 + 2x + x
2
.
Solution : We first arrange the terms of the dividend
and the divisor in the decreasing order of their degrees.
Arranging the terms in this order is termed as writing the
polynomials in its standard form. In this example, the
dividend is already in its standard form, and the divisor,
also in standard form, is x
2
+ 2x + 1.
Step 1 : To obtain the first term of the quotient, divide
the highest degree term of the dividend (i.e., 3x
3
) by the
highest degree term of the divisor (i.e., x
2
). This is 3x.
Then carry out the division process. What remains is –
5x
2
– x + 5.
Step 2 : Now, to obtain the second term of the quotient, divide the highest degree term of the
new dividend (i.e., 5x
2
) by the highest degree term of the divisor (i.e., x
2
). This gives – 5. Again
carry out the division process with– 5x
2
– x + 5.
Step 3 : What remains is 9x + 10. Now, the degree of 9x + 10 is less than the degree of the
divisor x
2
+ 2x + 1. So, we cannot continue the division any further.
So, the quotient is 3x – 5 and the remainder is 9x + 10. Also,
(x
2
+ 2x + 1) × (3x – 5) + (9x + 10)= (3x
3
+ 6x
2
+ 3x – 5x
2
– 10x – 5 + 9x + 10)
= 3x
3
+ x
2
+ 2x + 5
Here again, we see that
Dividend = Divisor × Quotient + Remainder
We are applying here an algorithm called Euclid’s division algorithm.
2
2
2 1
2 2 3 1
2 4
1
2
3
x
x x x
x x
x
x

  

 
 
 
 
2 3 2
3 2
2
2
3 5
2 1 3 2 5
3 6 3
5 5
5 10 5
9 10
x
x x x x x
x x x
x x
x x
x

    
 
  
  
  
  


71. Polynomials69
Free Distribution by A.P. Government
This says that
If p(x) and g(x) are any two polynomials with g(x)  0, then we can find polynomials
q(x) and r(x) such that
p(x) = g(x) × q(x) + r(x),
Where r(x) = 0 or degree of r(x) < degree of g(x).
This result is known as the Division Algorithm for polynomials.
Now, we have the following results from the above discussions
(i)If q(x) is linear polynomial then r(x) = r is a constant.
(ii)If degree of q(x) = 1, then degree of p(x) = 1 + degree of g(x).
(iii)If p(x) is divided by (x – a), then the remainder is p(a).
(iv)If r = 0, we say q(x) divides p(x) exactly or q(x) is a factor of p(x).
Let us now take some examples to illustrate its use.
Example-10. Divide 3x
2
– x
3
– 3x + 5 by x – 1 – x
2
, and verify the division algorithm.
Solution : Note that the given polynomials are not in standard form. To carry out division, we
first write both the dividend and divisor in decreasing orders of their degrees.
So,dividend = – x
3
+ 3x
2
– 3x + 5 and
divisor = – x
2
+ x – 1.
Division process is shown on the right side.
We stop here since degree of the remainder is less
than the degree of (–x
2
+ x – 1) the divisor.
So, quotient = x – 2, remainder = 3.
Now,
Dividend = Divisor × Quotient + Remainder
= (–x
2
+ x – 1) (x – 2) + 3
= – x
3
+ x
2
– x + 2x
2
– 2x + 2 + 3
= – x
3
+ 3x
2
– 3x + 5
In this way, the division algorithm is verified.
2 3 2
3 2
2
2
2
1 3 3 5
2 2 5
2 2 2
3
x
x x x x x
x x x
x x
x x

      
  
  
 
 
  

72. Mathematic70
Free Distribution by A.P. Government
Example-11. Find all the zeroes of 2x
4
– 3x
3
– 3x
2
+ 6x – 2, if you know that two of its zeroes
are 2 and –2.
Solution : Since two of the zeroes are 2 and – 2, therefore we can divide by (x –2 )
(x + 2) = x
2
– 2.
First term of quotient is
4
2
2
2
2
x
x
x

Second term of quotient is
3
2
3
3
x
x
x


Third term of quotient is
2
2
1
x
x

So, 2x
4
– 3x
3
– 3x
2
+ 6x – 2 = (x
2
– 2) (2x
2
– 3x + 1).
Now, by splitting –3x, we factorize 2x
2
– 3x + 1 as (2x – 1) (x – 1). So, its zeroes are
given by x =
1
2
and x = 1. Therefore, the zeroes of the given polynomial are 2, –2 and 1
and
1
2
.
EXERCISE – 3.4
1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in
each of the following :
(i)p(x) = x
3
– 3x
2
+ 5x – 3, g(x) = x
2
– 2
(ii)p(x) = x
4
– 3x
2
+ 4x + 5, g(x) = x
2
+ 1 – x
(iii)p(x) = x
4
– 5x + 6, g(x) = 2 – x
2
2
2 4 3 2
4 2
3 2
3
2
2
2 3 1
2 2 3 3 6 2
2 4
3 6 2
3 6
2
2
0
x x
x x x x x
x x
x x x
x x
x
x
 
    

 
   
 
 


 

73. Polynomials71
Free Distribution by A.P. Government
2. Check in which case the first polynomial is a factor of the second polynomial by dividing
the second polynomial by the first polynomial :
(i)t
2
– 3, 2t
4
+ 3t
3
– 2t
2
– 9t – 12
(ii)x
2
+ 3x + 1, 3x
4
+ 5x
3
– 7x
2
+ 2x + 2
(iii)x
3
– 3x + 1, x
5
– 4x
3
+ x
2
+ 3x + 1
3. Obtain all other zeroes of 3x
4
+ 6x
3
– 2x
2
– 10x – 5, if two of its zeroes are
5
3
and
5
3
 .
4. On dividing x
3
– 3x
2
+ x + 2 by a polynomial g(x), the quotient and remainder were
x – 2 and – 2x + 4, respectively. Find g(x).
5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm
and
(i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0
OPTIONAL EXERCISE
[This exercise is not meant for examination]
1. Verify that the numbers given alongside the cubic polynomials below are their zeroes.
Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x
3
+ x
2
– 5x + 2;
1
2
, 1, –2 (ii) x
3
+ 4x
2
+ 5x – 2; 1, 1, 1
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a
time, and the product of its zeroes as 2, –7, –14 respectively.
3. If the zeroes of the polynomial x
3
– 3x
2
+ x + 1 are a – b, a, a + b find a and b.
4. If two zeroes of the polynomial x
4
– 6x
3
– 26x
2
+ 138x – 35 are 2 3, find other
zeroes.
5. If the polynomial x
4
– 6x
3
– 16x
2
+ 25x + 10 is divided by another polynomial x
2
– 2x +
k, the remainder comes out to be x + a, find k and a.

74. Mathematic72
Free Distribution by A.P. Government
WHAT WE HAVE DISCUSSED
In this chapter, you have studied the following points:
1. Polynomials of degrees 1, 2 and 3 are called linear, quadratic and cubic polynomials
respectively.
2. A quadratic polynomial in x with real coefficients is of the form ax
2
+ bx + c, where a, b,
c are real numbers with a  0.
3. The zeroes of a polynomial p(x) are the x-coordinates of the points / where the graph of
y = p(x) intersects the x-axis.
4. A quadratic polynomial can have at most 2 zeroes and a cubic polynomial can have at
the most 3 zeroes.
5. If  and  are the zeroes of the quadratic polynomial ax
2
+ bx + c, then
 +  = –
b
a
, =
c
a
.
6. If , ,  are the zeroes of the cubic polynomial ax
3
+ bx
2
+ cx + d = 0, then
 +  +  =
b
a

,
 +  +  =
c
a
,
and =
d
a

.
7. The division algorithm states that given any polynomial p(x) and any non-zero polynomial
g(x), there are polynomials q(x) and r(x) such that
p(x) = g(x) q(x) + r(x),
where r(x) = 0 or degree r(x) < degree g(x).

76. 4.1 INTRODUCTION
One day Siri went to a book shop with her father and bought 3 notebooks and 2 pens. Her
father paid D80 for it. Her friend Laxmi liked the notebooks and pens so she bought 4 notebooks
and 3 pens of the same kind for D110.
Snigdha liked the pens and Madhu liked the notebooks. They asked Siri the cost of one
pen and one notebook. But, Siri did not know the cost of one notebook and one pen separately.
How can they find the costs?
In this example, the cost of a notebook and a pen are not known. These are unknown
quantities. We come across many such situations in our day-to-day life.
THINK AND DISCUSS
Two situations are given below:
(i)The cost of 1kg potatoes and 2kg tomatoes was D30 on a certain day. After two days, the
cost of 2kg potatoes and 4kg tomatoes was found to be D66.
(ii)The coach of a cricket team of M.K.Nagar High School buys 3 bats and 6 balls for D3900.
Later he buys one more bat and 2 balls for D1300.
Identify the unknowns in each situation. We observe that there are two unknowns in each
case.
4.1.1 HOW DO WE FIND UNKNOWN QUANTITIES?
In the introduction, Siri bought 3 notebooks and 2 pens for D80. How can we find the cost
of a notebook or the cost of a pen?
Snigdha and Madhu tried to guess. Snigdha said that price of each notebook could be D25.
Then three notebooks would cost D75, the two pens would cost D5 and each pen could be for
D2.50.
Madhu felt that D2.50 for one pen was too little. It should be at least D16. Then the price of
each notebook would also be D16.
We can see that there can be many possible values for the price of a notebook and of a pen
Pair of Linear Equations
in Two Variables4

77. Mathematic7 4
Free Distribution by A.P. Government
so that the total cost is D80. So, how do we find cost price at which Siri and Laxmi bought them?
By only using Siri's situation, we cannot find the costs. We have to use Laxmi's situation also.
4.1.2 USING BOTH EQ UATIONS TOGETHER
Laxmi also bought the same types of notebooks and pens as Siri. She paid D110 for 4
notebooks and 3 pens.
So, we have two situations which can be represented as follows:
(i)3 notebooks + 2 pens = D80.
(ii)4 notebooks + 3 pens = D110.
Does this help us find the cost of a pen and a notebook?
Consider the prices mentioned by Snigdha. If the price of one notebook is D25 and the
price of one pen is D2.50 then,
The cost of 4 notebooks would be:4 × 25 = D100
And the cost for 3 pens would be:3 × 2.50 = D7.50
If Snigdha is right then Laxmi should have paid D100 + 7.50 = D107.50 but she paid D110.
Now, consider the prices mentioned by Madhu. Then,
The cost of 4 notebooks, if one is for D16, would be: 4 × 16 = D64
And the cost for 3 pens, if one is for D16, would be: 3 × 16 = D 48
If Madhu is right then Laxmi should have paid D64 + 48 = D112 but this is more than the
price she paid.
So what do we do? How to find the exact cost of the notebook and the pen?
If we have only one equation but two unknowns (variables), we can find many solutions.
So, when we have two variables, we need at least two independent equations to get a unique
solution. One way to find the values of unknown quantities is by using the Model method. In this
method, rectangles or portions of rectangles are often used to represent the unknowns. Let us
look at the first situation using the model method:
Step-1 : Represent the number of notebooks and pens by blocks.
Siri bought 3 books and 2 pens for D80.
Laxmi bought 4 books and 3 pens for D110.
D80
D110

78. Pair of Linear Equations in Two Variables7 5
Free Distribution by A.P. Government
Step-2 : Increase (or decrease) the quantities in proportion to make one of the quantities equal in
both situations. Here, we make the number of pens equal.
In Step 2, we observe a simple proportional reasoning.
Since Siri bought 3 books and 2 pens for D80, so for 9 books and 6 pens:
3 × 3 = 9 books and 3 × 2 = 6 pens, the cost will be 3 × 80 = D240 (1)
Similarly, Laxmi bought 4 books and 3 pens for D110, so:
2 × 4 = 8 books and 2 × 3 = 6 pens will cost 2 × 110 = D220 (2)
After comparing (1) and (2), we can easily observe that 1 extra book costs
D240 - D 220 = D20. So one book is of D20.
Siri bought 3 books and 2 pens for D80. Since each book costs D20, 3 books cost D 60.
So the cost of 2 pens become D 80 - D 60 = D 20.
So, cost of each pen is D20 ÷ 2 = D10.
Let us try these costs in Laxmi's situation. 4 books will cost D 80 and three pens will cost
D 30 for a total of D 110, which is true.
From the above discussion and calculation, it is clear that to get exactly one solution (unique
solution) we need at least two independent linear equations in the same two variables.
In general, an equation of the form ax + by + c = 0 where a, b, c are real numbers and
where at least one of a or b is not zero, is called a linear equation in two variables x and y. [We
often write this condition as a
2
+ b
2
 0].
TRY THESE
Mark the correct option in the following questions:
1.Which of the following equations is not a linear equation?
a) 5 + 4x = y+ 3 b) x+2y = y - x
c) 3 - x = y
2
+ 4 d) x + y = 0
D240
(3×D80)
(3 books × 3) 9 books (2 pens × 3) 6 pens
D220
(2×D110)
(4 books × 2) 8 books (3 pens × 2) 6 pens

79. Mathematic7 6
Free Distribution by A.P. Government
2.Which of the following is a linear equation in one variable?
a) 2x + 1 = y - 3 b) 2t - 1 = 2t + 5
c) 2x - 1 = x
2
d) x
2
- x + 1 = 0
3.Which of the following numbers is a solution for the equation 2(x + 3) = 18?
a) 5 b) 6 c) 13 d) 21
4.The value of x which satisfies the equation 2x - (4 - x) = 5 - x is
a) 4.5 b) 3 c) 2.25d) 0.5
5.The equation x - 4y= 5 has
a) no solution b) unique solution
c) two solutions d) infinitely many solutions
4.2 SOLUTIONS OF PAIRS OF LINEAR EQUATIONS IN TWO VARIABLES
In the introductory example of notebooks and pens, how many equations did we have?
We had two equations or a pair of linear equations in two variables. What do we mean by the
solution for a pair of linear equations?
The pair of values of the variables x and y which together satisfy each one of the equations
is called the solution for a pair of linear equations.
4.2.1 GRAPHICAL METHOD OF FINDING SOLUTION OF A PAIR OF LINEAR EQUATIONS
What will be the number of solutions for a pair of linear equations in two variables? Is the
number of solutions infinite or unique or none?
In an earlier section, we used the model method for solving the pair of linear equations.
Now we will use graphs to solve the equations.
Let: a
1
x + b
1
y + c
1
= 0, (a
1
2
+ b
1
2
 0) and a
2
x + b
2
y + c
2
= 0 (a
2
2
+ b
2
2
 0) form a pair
of linear equation in two variables.
The graph of a linear equation in two variables is a straight line. Ordered pairs of real
numbers (x, y) representing points on the line are solutions of the equation and ordered pairs of
real numbers (x, y) that do not represent points on the line are not solutions.
When we have a pair of equations, they represent lines in the same plane. So, if we have
two lines in the same plane, what can be the possible relations between them? What is the
significance of this relation?

80. Pair of Linear Equations in Two Variables7 7
Free Distribution by A.P. Government
When two lines are drawn in the same plane, only one of the following three situations is
possible:
i)The two lines may intersect at one point.
ii)The two lines may not intersect i.e., they are parallel.
iii)The two lines may be coincident.
Let us write the equations in the first example in terms of x and y where x is the cost of a
notebook and y is the cost of a pen. Then, the equations are 3x + 6y = 80 and 4x + 3y = 110.
For the equation 3x + 6y = 80 For the equation 4x + 3y = 110
xy =
80 3
2
x
(x, y) xy =
110 4
3
x
(x, y)
0y =
80 3(0)
2

= 40(0, 40)-10y =
110 4( 10)
3
 
= 50(-10, 50)
10y =
80 3(10
2

= 25(10, 25)20y =
110 4(20)
3

= 10(20, 10)
20y =
80 3(20)
2

= 10(20, 10)50y =
110 4(50)
3

= -30(50, -30)
30y =
80 3(30
2

= -5(30, -5)
After plotting the above points in the Cartesian
plane, we observe that the two straight lines are
intersecting at the point (20, 10).
Substituting the values of x and y in equation
we get 3(20) + 2(10) = 80 and 4(20) + 3(10) = 110.
Thus, as determined by the graphical method,
the cost of each book is D20 and of each pen is D10.
Recall that we got the same solution using the model
method.
Since (20, 10) is the only common intersection point, there is only one solution for this pair
of linear equations in two variables. Such equations are known as consistent pairs of linear
equations. They will always have only a unique solution.
-90
Y
O
X
1
Y
1
X
-80-70-60-50-40-30-20-10102030405060708090
-20
-30
-40
-50
-60
-70
-80
-90
10
20
30
40
50
60
70
80
90
(20, 10)

81. Mathematic7 8
Free Distribution by A.P. Government
Now, let us look at the example from the think and discuss section. We want to find the
cost of 1kg of potatoes and the cost of 1 kg of tomatoes each. Let the cost of 1kg potatoes be
Dx and cost of 1kg of tomato be Dy. Then, the equations will become 1x+2y=30 and 2x+4y=66.
For the equation x + 2y = 30 For the equation 2x + 4y = 66
xy =
30
2
x
(x, y) xy =
66 2
4
x
(x, y)
0y =
30 0
2

= 15 (0, 15) 1y =
66 2(1)
4

= 16 (1, 16)
2y =
30 2
2

= 14 (2, 14) 3y =
66 2(3)
4

= 15(3, 15)
4y =
30 4
2

= 13 (4, 13) 5y =
66 2(5)
4

= 14(5, 14)
6y =
30 6
2

= 12 (6, 12) 7y =
66 2(7)
4

= 13(7, 13)
Here, we find that the situation is represented
graphically by two parallel lines. Since the lines do
not intersect at all, the equations have no common
solution. This means that the cost of the potato and
tomato was different on different days. We see this in
real life also. We cannot expect the same cost price
of vegetables every day; it keeps changing. Also, the
change is independent.
Such pairs of linear equations which have no
solution are known as inconsistent pairs of linear
equations.
In the second example from the think and
discuss section, let the cost of each bat be D x and each ball be D y. Then we can write the
equations as 3x + 2y = 3900 and x + 2y = 1300.
For the equation 3x + 2y = 3900 For the equation x + 2y = 1300
xy =
3900 3
6
x
(x, y) xy =
1300 3
2
x
(x, y)
100y =
3900 3(100)
6

= 600(100, 600)100y =
1300 100
2

= 600(100, 600)
-18
Y
O
X
1
Y
1
X
-16-14-12-10-8-6-42 24681012141618
-4
-6
-8
-10
-12
-14
-16
-18
2
4
6
8
10
12
14
16
18

82. Pair of Linear Equations in Two Variables7 9
Free Distribution by A.P. Government
200y =
3900 3(200)
6

= 550(200, 550)200y =
1300 200
2

= 550(200, 550)
300y =
3900 3(300)
6

= 500(300, 500)300y =
1300 300
2

= 500(300, 500)
400y =
3900 3(400)
6

= 450(400, 450)400y =
1300 400
2

= 450(400, 450)
We see that the equations are geometrically
shown by a pair of coincident lines. If the solutions
of the equations are given by the common points,
then what are the common points in this case?
From the graph, we observe that every point
on the line is a common solution to both the equations.
So, they have infinitely many solutions as both the
equations are equivalent. Such pairs of equations are
called dependent pair of linear equations in two
variables.
TRY THIS
In the example given above, can you find the
cost of each bat and ball?
THINK AND DISCUSS
Is a dependent pair of linear equations always consistent. Why or why not?
DO THESE
1.Solve the following systems of equations :
i)x - 2y = 0 ii)x + y = 2 iii)2x - y = 4
3x + 4y = 20 2x + 2y = 4 4x - 2y = 6
2.The two rails on a railway track are represented by the equations.
x + 2y - 4 = 0 and 2x + 4y- 12 = 0. Represent this situation graphically.
4.2.3 RELATION BETWEEN COEFFICIENTS AND NATURE OF SYSTEM OF EQUATIONS
Let a, b, c and a
2
, b
2
, c
2
denote the coefficients of a given pair of linear equations in two
variables. Then, let us write and compare the values of
1 1
2 2
,
a b
a b
and
1
2
c
c
in the above examples.
Y
O
X
1
Y
1
X
100200300400500600700800900
-
1
0
0
-
2
0
0
-
3
0
0
-
4
0
0
-
5
0
0
-
6
0
0
-
7
0
0
-
8
0
0
-
9
0
0
-200
-300
-400
-500
-600
-700
-800
-900
100
200
300
400
500
600
700
800
900

83. Mathematic8 0
Free Distribution by A.P. Government
Pair of lines
1
2
a
a
1
2
b
b
1
2
c
c
Comparison GraphicalAlgebraic
of ratios representationinterpretation
1.3x+2y–80=0
3
4
2
3
80
110


1 1
2 2
a b
a b

IntersectingUnique
4x+3y–110=0 lines solution
2.1x+2y–30=0
1
2
2
4
30
66


1 1 1
2 2 2
a b c
a b c
 
Parallel No solution
2x+3y–66=0 lines
3.3x+6y=3900
3
1
6
2
3900
1300
1 1 1
2 2 2
a b c
a b c
 
CoincidentInfinite
x+2y=1300 lines (Depen-number of
dent lines)solutions
Let us look at a few examples.
Example-1.Check whether the given pair of equations is for intersecting, parallel or coincident
lines. Find the solution if the equations are consistent.
Solution :2x + y - 5 = 0
3x - 2y - 4 = 0
1
2
2
3
a
a

1
2
1
2
b
b


1
2
5
4
c
c



Since
1 1
2 2
a b
a b

, therefore they are intersecting lines and hence, consistent pair of linear equation.
For the equation 2x + y = 5 For the equation 3x - 2y = 4
xy = 5 - 2x (x, y) xy =
4 3
2
x

(x, y0
0y = 5 - 2 (0) - 5(0, 5) 0y =
4 3(0)
2


= -2(0, -2)
1y = 5 - 2(1) = 3(1, 3) 2y =
4 3(2)
2


= 1 (2, 1)
2y = 5 - 2(2) = 1(2, 1) 4y =
4 3(4)
2


= 4 (4, 4)
3y = 5 - 2(3) = 1(3, -1)
4y = 5 - 2(4) = 3(4, -3)

84. Pair of Linear Equations in Two Variables8 1
Free Distribution by A.P. Government
The unique solution of this pair of equations is (2,1).
Example-2.Check whether the following pair of equations is consistent.
3x + 4y = 2 and 6x + 8y = 4
Verify by drawing a graphical representation.
Solution :3x + 4y - 2 = 0
6x+8y - 4 = 0
1
2
3 1
6 2
a
a
 
1
2
4 1
8 2
b
b
 
1
2
2 1
4 2
c
c

 

Since
1 1 1
2 2 2
a b c
a b c
 
, therefore, they are coincident lines. So, the pair of linear equations
is not consistent.
For the equation 3x + 4y = 2 For the equation 6x - 8y = 4
xy =
2 3
4
x
(x, y) xy =
4 6
8
x
(x, y)
0y =
4 3(0)
4

=
1
2
(0,
1
2
) 0y =
4 6(0)
8

=
1
2
(0,
1
2
)
2y =
4 3(2)
4

= -1(2, -1) 2y =
4 6(2)
8

= -1(2, -1)
4y =
4 3(4)
4

= -2.5(4, -2.5)4y =
4 6(4)
8

= -2.5(4, -2.5)
6y =
4 3(6)
4

= -4(6, -4) 6y =
4 6(6)
8

= -4(6, -4)
-9
Y
O
X
1
Y
1
X
-8-7-6-5-4-3-2-1 123456789
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
(2,1)

85. Mathematic8 2
Free Distribution by A.P. Government
Example-3.Check whether the equations2x-3y = 5 and 4x-6y = 9 are consistent. Also show
graphical representation.
Solution : 4x-6y - 9 = 0
2x-3y - 5 = 0
1
2
4 2
2 1
a
a
 
1
2
6 2
3 1
b
b

 

1
2
9 9
5 5
c
c

 

1 1 1
2 2 2
a b c
a b c
 
So the equations are inconsistent. They have no solutions and its graph is of parallel lines.
For the equation 4x - 6y = 9 For the equation 2x - 3y = 5
xy =
9 4
6
x

(x, y) xy =
5 2
3
x

(x, y)
0y =
9 0 3
6 2
 


(0, -1.5)0y =
5 2(0) 5
3 3
 


(0, -1.5)
3y =
3 1
6 2



(3, 0.5)3y =
5 2(3) 1
3 3



(3, 0.3)
6y =
15 5
6 2
 


(6, 2.5)6y =
7 7
3 3



(6, 2.3)
-9
Y
O
X
1
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86. Pair of Linear Equations in Two Variables8 3
Free Distribution by A.P. Government
DO THIS
Check each of the given systems of equations to see if it has a unique solution, infinitely
many solutions or no solution. Solve them graphically.
(i)2x+3y = 1 (ii)x + 2y = 6 (iii)3x - 2y = 6
3x-y = 7 2x + 4y = 12 6x+4y = 18
TRY THESE
1.For what value of 'p' the following pair of equations has a unique solution.
2x + py = - 5 and 3x + 3y = - 6
2.Find the value of 'k' for which the pair of equations 2x - ky + 3 = 0, 4x + 6y - 5 =0
represent parallel lines.
3.For what value of 'k', the pair of equation 3x + 4y + 2 = 0 and 9x + 12y + k = 0 represent
coincident lines.
4.For what values of 'p' will the following pair of liner equation have infinitely many solutions?
px + 3y - (p - 3) = 0
12x + py - p = 0
Let us look at some more examples.
Example-4.In a garden there are some bees and flowers. If one bee sits on each flower then
one bee will be left. If two bees sit on each flower, one flower will be left. Find the number of
bees and number of flowers.
Solution :Let the number of bees = x and
the number of flowers = y
If one bee sits on each flower then one bee will be left. So, x = y + 1
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87. Mathematic8 4
Free Distribution by A.P. Government
or x - y - 1 = 0 (1)
If two bees sit on each flower, one flower will be left. So, x = 2(y- 1)
or x - 2y+2 = 0 (2)
For the equation x - y - 1 = 0 For the equation x - 2y + 2 = 0
xy = x - 1 (x, y) xy =
2
2
x
(x, y)
0y = 0 - 1 = -1(0, -1) 0y =
0 2
2

= 1 (0, 1)
1y = 1 - 1 = 0 (1, 0) 2y =
2 2
2

= 2 (2, 2)
2y = 2 - 1 = 1 (2, 1) 4y =
4 2
2

= 3 (4, 3)
3y = 3 - 1 = 2 (3, 2) 6y =
6 2
2

= 4 (6, 4)
4y = 4 - 1 = 3 (4, 3)
Therefore, there are 4 bees and 3 flowers.
Example-5.The perimeter of a rectangular plot of land is 32m. If the length is increased by 2m
and the breadth is decreased by 1m, the area of the plot remains the same. Find the length and
breadth of the plot.
Solution : Let length and breadth of the rectangular land be l and b respectively. Then,
area = lb and
Perimeter = 2(l + b) = 32 m.
l + b = 16or l + b - 16 = 0 (1)
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(4, 3)

88. Pair of Linear Equations in Two Variables8 5
Free Distribution by A.P. Government
When length is increased by 2 m., then new length is l + 2. Also breadth is decreased by
1m so new breadth is b - 1.
Then,area= (l + 2) (b - 1)
Since there is no change in the area, therefore
(l + 2) (b - 1) = lb
lb - l +2b - 2 = lb or lb - lb = 1 - 2b + 2
l - 2b + 2 = 0 (2)
For the equation l + b - 16 = 0 For the equation l - 2b + 2 = 0
xb = 16 - l (l, b) xb =
1 2
2

(l, b)
6b = 16 - 6 = 10(6, 10)6b =
6 2
2

= 4 (6, 4)
8b = 16 - 8 = 8 (8, 8) 8b =
8 2
2

= 5 (8, 5)
10b = 16 - 10 = 6(10, 6)10b =
10 2
2

= 6 (10, 6)
12b = 16 - 12 = 4(12, 4)12b =
12 2
2

= 7 (12, 7)
14b = 16 - 14 = 2(14, 2)14b =
14 2
2

= 8 (14, 8)
So, original length of the plot is 10m and its breadth is 6m.
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89. Mathematic8 6
Free Distribution by A.P. Government
EXERCISE - 4.1
1.By comparing the ratios
1 1 1
2 2 2
, ,
a b c
a b c
and without drawing graphs, find out whether the
lines represented by the following pairs of linear equations intersect at a point, are parallel
or are coincident.
a)5x- 4y + 8 = 0 b)9x+3y + 12 = 0 c)6x - 3y + 10 = 0
7x+6y-9 = 0 18x+6y + 24 = 0 2x - y + 9 = 0
2.Check whether the following equations are consistent or inconsistent. Solve them graphically.
a)3x+2y=5 b)2x - 3y = 8 c)
3 5
2 3
x y = 7
2x - 3y=7 4x - 6y = 9 9x - 10y=14
d)5x-3y = 11 e)
4
3
x+2y = 8 f)x + y = 5
-10x+6y = -22 2x+3y = 12 2x+2y = 10
g)x - y = 8 h)2x + y-6 = 0 i)2x-2y - 2 = 0
3x-3y = 16 4x-2y- 4 = 0 4x-4y- 5 = 0
3.Neha went to a 'sale' to purchase some pants and skirts. When her friend asked her how
many of each she had bought, she answered "The number of skirts are two less than twice
the number of pants purchased. Also the number of skirts is four less than four times the
number of pants purchased."
Help her friend find how many pants and shirts Neha bought.
4.10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more
than the number of boys then, find the number of boys and the number of girls who took
part in the quiz.
5.5 pencils and 7 pens together cost D50 whereas 7 pencils and 5 pens together cost D46.
Find the cost of one pencil and that of one pen.
6.Half the perimeter of a rectangular garden, whose length is 4m more than its width, is 36m.
Find the dimensions of the garden.
7.We have a linear equation 2x + 3y - 8 = 0. Write another linear equation in two variables
such that the geometrical representation of the pair so formed is intersecting lines.
Now, write two more linear equations so that one forms a pair of parallel lines and the
second forms coincident line with the given equation.
8.The area of a rectangle gets reduced by 80 sq units it its length is reduced by 5 units and
breadth is increased by 2 units. If we increase the length by 10 units and decrease the

90. Pair of Linear Equations in Two Variables8 7
Free Distribution by A.P. Government
breadth by 5 units, the area will increase by 50 sq units. Find the length and breadth of the
rectangle.
9.In class 9, if three students sit on each bench, one student will be left. If four students sit on
each bench, one bench will be left. Find the number of students and the number of benches
in that class.
4.3 ALGEBRAIC METHODS OF FINDING THE SOLUTIONS FOR A PAIR OF LINEAR
EQUATIONS
We have learnt how to solve a pair of linear equations graphically. But, the graphical method
is not convenient in cases where the point representing the solution has no integral co-ordinates.
For example, when the solution is of the form (3, 27), (- 1.75, 3.3), (
4
13
,
1
19
) etc. There
is every possibility of making mistakes while reading such co-ordinates. Is there any alternative
method of finding the solution? There are several algebraic methods, which we shall discuss now.
4.3.1 SUBSTITUTI ON METHOD
This method is useful for solving a pair of linear equations in two variables where one
variable can easily be written in terms of the other variable. To understand this method, let us
consider it step-wise
Step-1 : In one of the equations, express one variable in terms of the other variable. Say y in
terms of x.
Step-2 :Substitute the value of y obtained in step 1 in the second equation.
Step-3 :Simplify the equation obtained in step 2 and find the value of x.
Step-4 :Substitute the value of x obtained in step 3 in either of the equations and solve it for y.
Step-5 :Check the obtained solution by substituting the values of x and y in both the original
equations.
Example-6.Solve the given pair of equations using substitution method.
2x - y = 5
3x + 2y = 11
Solution :2x - y = 5 (1)
3x + 2y = 11 (2)
Equation (1) can be written as
y = 2x - 5 (Step 1)
Substituting in equation (2) we get
3x + 2(2x - 5) = 11 (Step 2)

91. Mathematic8 8
Free Distribution by A.P. Government
3x + 4x - 10 = 11
7x = 11 + 10 = 21
x = 21/7 = 3. (Step 3)
Substitute x =3 in equation (1) or (2)
2(3) - y = 5 (Step 4)
y = 6 - 5 = 1
Therefore, required solution is x = 3 and y = 1.
Both the equations are satisfied by x = 3 and y = 1. (Step 5)
DO THESE
Solve each pair of equation by using the substitution method.
1)3x - 5y = -1 2)x+2y = - 1 3)2x+3y = 9
x - y = - 1 2x - 3y = 12 3x+4y = 5
4)
6
x
y

= 6 5)0.2x + 0.3y = 13 6)2 + 3 = 0x y
8
3x
y

= 5 0.4x + 0.5y = 2.3 3 - 8 = 0x y
4.3.2 ELIMINATION METHOD
In this method, first we eliminate (remove) one of the two variables by equating its coefficients.
This gives a single equation which can be solved to get the value of the other variable. To understand
this method, let us consider it stepwise.
Step-1 :Write both the equations in the form of ax + by = c.
Step-2 :Make the coefficients of one of the variables, say 'x', numerically equal by multiplying
each equation by suitable real numbers.
Step-3 :If the variable to be eliminated has the same sign in both equations, subtract the two
equations to get an equation in one variable. If they have opposite signs then add.
Step-4 :Solve the equation for the remaining variable.
Step-5 :Substitute the value of this variable in any one of the original equations and find the
value of the eliminated variable.
Example-7.Solve the following pair of linear equations using elimination method.
3x + 2y = 11
2x + 3y = 4

92. Pair of Linear Equations in Two Variables8 9
Free Distribution by A.P. Government
Solution :3x + 2y = 11 (1)
2x + 3y = 4 (2) (Step 1)
Let us eliminate 'y' from the given equations. The coefficients of 'y' in the given equations are 2
and 3. L.C.M.of 2 and 3 is 6. So, multiply equation (1) by 3 and equation (2) by 2.
Equation (1) × 39x + 6y = 33 (Step 2)
Equation (2) × 24x + 6y = 8
(-) (-) (-) (Step 3)
5x = 25
x =
25
5
= 5 (Step 4)
Substitute x = 5, in equation (1)
3(5) + 2y = 11
2y = 11 - 15 = - 4 (Step 5)
Therefore, the required solution is x = 5, y = - 2.
DO THESE
Solve each of the following pairs of equations by using the elimination method.
1.8x+ 5y = 9 2.2x + 3y = 8 3.3x + 4y = 25
3x+2y = 4 4x + 6y = 7 5x - 6y = -9
TRY THIS
Solve the given pair of linear equations
(a - b)
x
+ (a + b)
y
= a
2
- 2ab - b
2
(a + b) (x + y) = a
2
+ b
2
Some more examples:
Example-8.Snigdha went to a bank to withdraw D2000. She asked the cashier to give D50
and D100 notes only. Snigdha got 25 notes in all. Can you tell how many notes of D50 and D100
she received?
Solution :Let the number of D50 notes be x;
Let the number of D100 notes be y;
then,x + y = 25 (1)
50x + 100y = 2000 (2)
Kavitha used the substitution method.

93. Mathematic9 0
Free Distribution by A.P. Government
From equation (1)) x = 25 -y
Substituting in equation (2) 50 (25 - y) + 100y = 2000
1250 - 50y + 100y = 2000
50y = 2000 - 1250 = 750
y =
750
50
= 15
x = 25 - 15 = 10
Hence, Snigdha received ten D50 notes and fifteen D100 notes.
Laxmi used the elimination method to get the solution.
In the equations, coefficients of x are 1 and 50 respectively. So,
Equation (1) × 5050x + 50y = 1250
Equation (2) × 150x + 100y = 2000same sign, so subtract the equations
(-) (-) (-)
- 5y= -750
Or y =
750
50


= 15
Substitute y in equation (1)x + 15 = 25
x = 25 - 15 = 10
Hence Snigdha received ten D50 notes and fifteen D100 rupee notes.
Example-9.In a competitive exam, 3 marks are be awarded for every correct answer and for
every wrong answer, 1 mark is deducted. Madhu scored 40 marks in this exam. Had 4 marks
been awarded for each correct answer and 2 marks deducted for each incorrect answer, Madhu
would have scored 50 marks. How many questions were there in the test? (Madhu attempted all
the questions)
Solution :Let the number of correct answers be x;
and the number of wrong answers be y.
When 3 marks are given for each correct answer and 1 mark deducted for each wrong
answer, his score is 40 marks.
3x - y = 40 (1)
His score would have been 50 marks if 4 marks were given for each correct answer and 2
marks deducted for each wrong answer.
4x - 2y = 50 (2)

94. Pair of Linear Equations in Two Variables9 1
Free Distribution by A.P. Government
Substitution method
From equation (1), y = 3x - 40
Substitute in equation (2) 4x - 2 (3x - 40) = 50
4x - 6x + 80 = 50
- 6x = 50 - 80 = -30
x =
30
2


=15
Substitute the value of x in equation (1)
3(15) - y = 40
45 - y = 40
y = 45 - 40 = 5
DO THIS
Now use the elimination method to solve the above example-9.
Example-10. Mary told her daughter, "Seven years ago, I was seven times as old as you are
now. Also, three years from now, I shall be three times as old as you will be." Find the present
age of Mary and her daughter.
Solution :Let Mary's present age be x years and her daughter's age be y years.
Then, seven years ago Mary's age was x - 7 and daughter's age was y - 7.
x - 7 = 7(y - 7)
x - 7 = 7y - 49
x - 7y + 42 = 0 (1)
Three years hence, Mary's age will be x + 3 and daughter's age will be y + 3.
x + 3 = 3 (y + 3)
x + 3 = 3y + 9
x - 3y - 6 = 0 (2)
Elimination method
Equation 1 x - 7y = - 42
Equation 2 x - 3y = 6
(-) (+) (-)same sign for x, so subtract the two equations.
-4y = -48

95. Mathematic9 2
Free Distribution by A.P. Government
y =
48
4


= 12
Substitute the value of y in equation (2)
x-3 (12) - 6 = 0
x = 36 + 6 = 42
Therefore, Mary's age now is 42 years and her daughter's age is 12 years.`
DO THIS
Solve example-10 by using the substitution method.
Example-11. A publisher is planning to produce a new
textbook. The fixed costs (reviewing, editing, typesetting
and so on) are D 31.25 per book. In addition, he also
spends another D 320000 in producing the book. The
wholesale price (the amount received by the publisher) is
D 43.75 per book. How many books must the publisher
sell to break even, i.e., so that the costs will equal revenues?
Solution : The publisher breaks even when costs equal revenues. If x represents the number of
books printed and sold and y be the breakeven point, then the cost and revenue equations for the
publisher are
Cost equation is given by y = 320000 + 31.25x (1)
Revenue equation is given by y = 43.75x (2)
Using the second equation to substitute for y in the first equation, we have
43.75x = 3,20,000 + 31.25x
12.5x = 3,20,000
x =
3,20,000
12.5
= 25,600
Thus, the publisher will break even when 25,600 books are printed and sold.
EXERCISE - 4.2
Form a pair of linear equations for each of the following problems and find their solution.
1.The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If
each of them manages to save D2000 per month, find their monthly income.
2.The sum of a two digit number and the number obtained by reversing the digits is 66. If the
digits of the number differ by 2, find the number. How many such numbers are there?
The point which corresponds to
how much money you have to earn
through sales in order to equal the
money you spent in production is
break even point.

96. Pair of Linear Equations in Two Variables9 3
Free Distribution by A.P. Government
3.The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
4.The taxi charges in Hyderabad consist of a fixed charge, along with the charge for the
distance covered. For a distance of 10 km., the charge paid is D220. For a journey of
15 km. the charge paid is D310.
i.What are the fixed charges and charge per km?
ii.How much does a person have to pay for travelling a distance of 25 km?
5.A fraction becomes
4
5
if 1 is added to both numerator and denominator. If, however, 5 is
subtracted from both numerator and denominator, the fraction becomes
1
2
. What is the
fraction?
6.Places A and B are 100 km apart on a highway. One car starts from A and another from B
at the same time at different speeds. If the cars travel in the same direction, they meet in 5
hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the
two cars?
7.Two angles are complementary. The larger angle is 3° less than twice the measure of the
smaller angle. Find the measure of each angle.
8.An algebra textbook has a total of 1382 pages. It is broken up into two parts. The second
part of the book has 64 pages more than the first part. How many pages are in each part
of the book?
9.A chemist has two solutions of hydrochloric acid in stock. One is 50% solution and the
other is 80% solution. How much of each should be used to obtain 100ml of a 68%
solution.
10.Suppose you have D12000 to invest. You have to invest some amount at 10% and the rest
at 15%. How much should be invested at each rate to yield 12% on the total amount
invested?
4.4 EQUATIONS REDUCIBLE TO A PAIR OF LINEAR EQUATIONS IN TWO
VARIABLES
Now we shall discuss the solution of pairs of equations which are not linear but can be
reduced to linear form by making suitable substitutions. Let us see an example:
Example-12. Solve the given pair of equations.
2 3
x y

= 13
5 4
x y

= -2
Solution : Observe the given pair of equations. They are not linear equations. (Why?)

97. Mathematic9 4
Free Distribution by A.P. Government
We have2
1
x
 
 
 
+ 3
1
y
 
 
 
= 13 (1)
5
1
x
 
 
 
- 4
1
y
 
 
 
= -2 (2)
If we substitute
1
x
= p and
1
y
= q, we get a pair of linear equations:
2p + 3q = 13 (3)
5p - 4q = -2 (4)
Coefficients of q are 3 and 4 and their l.c.m. is 12. Using the elimination method:
Equation (3) × 48p + 12q = 52
Equation (4) × 315p - 12q = -6 'q' has opposite sign, so we add the two equations.
23p= 46
p =
46
23
= 2
Substitute the value of p in equation (3)
2(2) +3q = 13
3q = 13 - 4 = 9
q =
9
3
= 3
But,
1
x
= p = 2 x =
1
2
1
y
= q = 3 y =
1
3
Example-13. Kavitha thought of constructing 2 more rooms in her house. She enquired
about the labour. She came to know that 6 men and 8 women could finish this work in 14 days.
But she wanted the work completed in only 10 days. When she enquired, she was told that 8
men and 12 women could finish the work in 10 days. She wondered how much time would be
taken to finish the work if one man or one woman worked alone?
Solution :Let the time taken by one man to finish the work = x days.

98. Pair of Linear Equations in Two Variables9 5
Free Distribution by A.P. Government
Work done by one man in one day =
1
x
Let the time taken by one woman to finish the work= y days.
Work done by one woman in one day =
1
y
Now, 8 men and 12 women can finish the work in 10 days.
So work done by 8 men and 12 women in one day=
1
10
(1)
Also, work done by 8 men in one day is 8 ×
1
x
. =
8
x
Similarly, work done by 12 women in one day is 12 ×
1
y
=
12
y
Total work done by 8 men and 12 women in one day=
8 12
x y

(2)
Equating equations (1) and (2)
8 12 1
10x y
 
 
 
 
10
8 12
x y
 

 
 
= 1
80 120
x y

= 1 (3)
Also, 6 men and 8 women can finish the work in 14 days.
Work done by 6 men and 8 women in one day =
6 8 1
14x y
 
 14
6 8
x y
 

 
 
= 1
84 112
x y
 

 
 
(4)

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Free Distribution by A.P. Government
Observe equations (3) and (4). Are they linear equations? How do we solve them then? We can
convert them into linear equations by substituting
1
x
= u and
1
y
= v.
Equation (3) becomes 80u + 120v = 1 (5)
Equation (4) becomes 84u + 112v = 1 (6)
L.C.M. of 80 and 84 is 1680. Using the elimination method,
Equation (3) × 21 (21 × 80)u + (21 × 120)v = 21
Equation (4) × 20 (20 × 84)u + (20 × 112)v = 20
1680u+2520v = 21
1680u+2240v = 20 Same sign for u, so subtract
(-) (-) (-)
280v = 1
v =
1
280
Substitute in equation (5)80u + 120 ×
1
280
= 1
80u = 1
3
7
=
7 3
7

=
4
7
1
20
4 1 1
= =
7 80 140
u 
So one man alone can finish the work in 140 days and one woman alone can finish the
work in 280 days.
Example-14. A man travels 370 km partly by train and partly by car. If he covers 250 km by
train and the rest by car, it takes him 4 hours. But if he travels 130 km by train and the rest by car,
it takes 18 minutes more. Find the speed of the train and that of the car.
Solution :Let the speed of the train be x km. per hour and that of the car be y km. per hour.
Also, we know that time =
Distance
Speed
In situation 1, time spent travelling by train=
250
x
hrs.
And time spent travelling by car=
120
y
hrs.

100. Pair of Linear Equations in Two Variables9 7
Free Distribution by A.P. Government
So, total time taken = time spent in train + time spent in car =
250 120
+
x y
But, total time of journey is 4 hours, so
250 120
+ = 4
x y
125 60
+ = 2
x y
 (1)
Again, when he travels 130 km by train and the rest by car
Time taken by him to travel 130 km by car =
130
x
hrs.
Time taken by him to travel 240 km (370-310) by train =
240
y
hrs.
Total time taken =
130 240
+
x y
But given, time of journey is 4 hrs 18 min i.e.,
3
10
18
4
60
hrs.
So,
130 240 43
+ =
10x y
(2)
Substitute
1
x
= a and
1
y
= b in equations (1) and (2)
125a + 60b = 2 (3)
130a+ 240b = 43/10 (4)
For 60 and 240, l.c.m. is 240. Using the elimination method,
Equation (3) × 4500a+240b= 8
Equation (4) × 1130a+240b =
43
10
(Same sign, so subtract)
(-) (-) (-)
370a = 8 -
43 80 43 37
= =
10 10 10


101. Mathematic9 8
Free Distribution by A.P. Government
a =
10
37 1 1
=
10 370 100

Substitute a =
1
100
in equation (3)
5
4
1
125
100
 
 
 
 
+ 60b = 2
60b = 2 -
5 8 5 3
= =
4 4 4

b =
20
3 1 1
=
4 60 80

So a =
1
100
and b =
1
80
So
1 1
=
100x
and
1 1
=
80y
x = 100 km/hr and y = 80 km/hr.
So, speed of train was 100 km/hr and speed of car was 80 km/hr.
EXERCISE - 4.3
Solve each of the following pairs of equations by reducing them to a pair of linear equations.
i)
5 1
+
1 2x y 
= 2 ii)
= 2
x y x y
xy xy
 
= 6
6 3
-
1 2x y 
= 1
iii)
2 3
+
x y
= 2 iv)6x+3y = 6xy
4 9
-
x y
= 1 2x + 4y = 5xy
v)
5 2
-
x y x y 
= 1 vi)
2 3
+
x y
= 13

102. Pair of Linear Equations in Two Variables9 9
Free Distribution by A.P. Government
15 7
+
x y x y 
= 10 where x0, y0
5 4
-
x y
= -2 where x0, y0
vii)
10 2
+
x y x y 
= 4 viii)
1 1 3
+ =
3 3 4x y x y 
15 5
-
x y x y 
= -2
 
1 1 1
- =
2 3 2(3 ) 8x y x y

 
2.Formulate the following problems as a pair of equations and then find their solutions.
i.A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can
go 40 km upstream and 55 km downstream. Determine the speed of the stream and
that of the boat in still water.
ii.Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours
if he travels 120 km by train and rest by car. He takes 20 minutes more if he travels
200 km by train and rest by car. Find the speed of the train and the car.
iii.2 women and 5 men can together finish an embroidery work in 4 days while 3 women
and 6 men can finish it in 3 days. Find the time taken by 1 woman alone and 1 man
alone to finish the work.
OPTIONAL EXERCISE
[This exercise is not meant for examination]
1.Solve the following equations:-
(i)
2
+
x y
a b
= 2 (ii)
1 1
+
2 3
x y 
= 8
-
x y
a b
= 4
1 1
+
3 2
x y 
= 9
(iii) +
7 3
x y
= 5 (iv)3 + 2 = 3x y
-
2 9
x y
= 6 5 + 3 = 3x y
(v) -
ax by
b a
= a + b (vi)2
x
+ 3
y
= 17
ax - by = 2ab 2
x+2
+ 3
y+1
= 5

103. Mathematic1 0 0
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2.Animals in an experiment are to be kept on a strict diet. Each animal is to receive among
other things 20g of protein and 6g of fat. The laboratory technicians purchased two food
mixes, A and B. Mix A has 10% protein and 6% fat. Mix B has 20% protein and 2% fat.
How many grams of each mix should be used?
WHAT WE HAVE DISCUSSED
1.Two linear equations in the same two variables are called a pair of linear equations in two
variables.
a
1
x + b
1
y + c
1
= 0 (a
1
2
+ b
1
2
 0)
a
2
x + b
2
y + c
2
= 0 (a
2
2
+ b
2
2
 0)
Where a
1
, a
2
, b
1
, b
2
, c
1
, c
2
are real numbers.
2.A pair of linear equations in two variables can be solved using various methods.
3.The graph of a pair of linear equations in two variables is represented by two lines.
i.If the lines intersect at a point then the point gives the unique solution of the two equations.
In this case, the pair of equations is consistent.
ii.If the lines coincide, then there are infinitely many solutions - each point on the line
being a solution. In this case, the pair of equations is dependent.
iii.If the lines are parallel then the pair of equations has no solution. In this case, the pair
of equations is inconsistent.
4.We have discussed the following methods for finding the solution(s) of a pair of linear
equations.
i.Model Method.
ii.Graphical Method
iii.Algebraic methods - Substitution method and Elimination method.
5.There exists a relation between the coefficients and nature of system of equations.
i.If
1 1
2 2

a b
a b

, the pair of linear equations is consistent.
ii.If
1 1 1
2 2 2
-
a b c
a b c

, the pair of linear equations is inconsistent.
iii.If
1 1 1
2 2 2
= =
a b c
a b c
, the pair of linear equations is dependent and consistent.
6.There are several situations which can be mathematically represented by two equations
that are not linear to start with. But we can alter them so that they are reduced to a pair of
linear equations.

104. 5 . 1INTRO DUCTIO N
School sports committee wants to construct a Kho-Kho court of dimension 29 m. × 16 m.
This is to be a rectangular enclosure of
area 800 m
2
. They want to lease a space
of equal width all around the court for
the spectators. What would be the width
of the space for spectators? Would it be
enough?
Suppose the width of the space be
x meter. So from the figure length of the
plot would be (29 + 2x) meter.
And, breath of the rectangular plot would be= (16 + 2x) m.
Therefore, area of the rectangular plot will be= length × breadth
= (29 + 2x) (16 + 2x)
Since the area of the plot is = 800 m
2
(29 + 2x) (16 + 2x) = 800
4x
2
+ 90x + 464 = 800
4x
2
+ 90x - 336 = 0 (dividing by 2)
2x
2
+ 45x - 168 = 0
2x
2
+ 45 x - 168 = 0 ..... (1)
We solved the linear equations to find the value of ‘x’. Similarly, the value of x from the
above equation will give the possible width of the space for spectators.
Can you think of more such examples where we have to find the quantities like in above
example and get such equations.
Let us consider another example:
Rajni has a square metal sheet. She removed squares of side 9 cm. from each corner of this
sheet. Of the remaining sheet, she turned up the sides to form an open box as shown. The
capacity of the box is 144 cc. Can we find out the dimensions of the metal sheet?
Quadratic Equations5
29 .m
1
6

.
m
x
x
29+2 .x m
1
6
+
2

.
x
m

105. Mathematic1 0 2
Free Distribution by A.P. Government
Suppose the side of the square piece of metal sheet
be ‘x’ cm.
Then, the dimensions of the box are
9 cm. × (x-18) cm. × (x-18) cm.
Since volume of the box is 144 cc
9 (x-18) (x-18) = 144
(x-18)
2
= 16
x
2
- 36x + 308 = 0
So, the side ‘x’ of the metal sheet will satisfy the
equation.
x
2
- 36x + 308 = 0 ..... (2)
Are they quadratic polynominals?
We studied such quadratic polynomials of the form ax
2
+ bx + c, a
 0 in the previous chapter.
Since, the LHS of the above equations are quadratic polynomials they area called quadratic
equations.
In this chapter we will study quadratic and methods to find their roots.
5.2 QUADRATIC EQUATIONS
A quadratic equation in the variable x is an equation of the form ax
2
+ bx + c = 0, where
a, b, c are real number a  0. For example, 2x
2
+ x  300 = 0 is quadratic equation, Similarly,,
2x
2
 3x + 1 = 0, 4x  3x
2
+ 2 = 0 and 1  x
2
+ 300 = 0 are also quadratic equations.
In fact, any equation of the form p(x)  0, where p(x) is polynomial of degree 2, is a
quadratic equation. But when we write the terms of p(x) in descending order of their degrees,
then we get the standard form of the equation. That is, ax
2
+ bx + c = 0, a  0 is called the
standard form of a quadratic equation.
TRY THESE
Check whether the following equations are quadratic or not ?
(i)x
2
 6x  4 = 0 (ii)x
3
 6x
2
+ 2x  1 = 0
(iii)7x = 2x
2
(iv)
2
2
1
2x
x
 
(v)(2x + 1) (3x + 1) = b(x  1) (x  2)(vi)3y
2
= 192
9 cm.9 cm.
9 cm. 9 cm.
x cm.
x
c
m
.
9 cm.
x - 18
x
- 1
8

106. Quadratic Equations1 0 3
Free Distribution by A.P. Government
There are various uses of Quadratic Equations some of them are:-
1.When the rocket is fired upward, then the height of the rocket is defined
by a ‘quadratic equation.’
2.Shapes of the satellite dish, reflecting mirror in a telescope, lens of the
eye glasses and orbits of the celestial objects are defined by the qradratic
equations.
Satellite Dish Reflecting MirrorLens of Eye Glasses
3.The path of a projectile is defined by quadratic equation.
4.When the breaks are applied to a vehicle, the stopping
distance is calculated by using quadratic equation.
Example-1.Represent the following situations mathematically:
i.John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the
product of the number of marbles now they have is 124 marbles. We would like to find
out how many marbles they had previously.
ii.The hypotenuse of a right triangle is 25 cm. We know that the difference in lengths of
the other two sides is 5 cm. We would like to find out the length of the two sides?
Solution :i.Let the number of marbles John had be x.
Earth
Sun
Earth
Artificial
Satellite

107. Mathematic1 0 4
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Then the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = (45 – x) – 5
= 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x
2
– 200 + 5x
= – x
2
+ 45x – 200
So,– x
2
+ 45x – 200 = 124 (Given that product = 124)
i.e.,– x
2
+ 45x – 324 = 0
i.e., x
2
– 45x + 324 = 0 (Multiply -ve sign)
Therefore, the number of marbles John had ‘x’, satisfies the quadratic equation
x
2
– 45x + 324 = 0
which is the required representation of the problem mathematically.
ii.We know that (hypotenuse)
2
= (adjacent side)
2
+ (opposite side)
2
Let the length of smaller side be x cm.
Then length of larger side = (x + 5) cm.
Given length of hypotenuse = 25 cm.
So,x
2
+ (x + 5)
2
= (25)
2
x
2
+ x
2
+ 10x + 25= 625
2x
2
+ 10x - 600= 0
x
2
+ 5x - 300 = 0
Value of x from the above equation will give the possible value of length of sides of the
given right angled triangle.
Example-2.Check whether the following are quadratic equations:
i.(x – 2)
2
+ 1 = 2x – 3 ii.x(x + 1) + 8 = (x + 2) (x – 2)
iii.x (2x + 3) = x
2
+ 1 iv.(x + 2)
3
= x
3
– 4
Solution :i.LHS = (x – 2)
2
+ 1 = x
2
– 4x + 4 + 1 = x
2
– 4x + 5
Therefore, (x – 2) 2 + 1 = 2x – 3 can be written as
x
2
– 4x + 5 = 2x – 3
x
x
c
m
+
5

.
2
5

.
c
m

108. Quadratic Equations1 0 5
Free Distribution by A.P. Government
i.e.,x
2
– 6x + 8 = 0
It is in the form of ax
2
+ bx + c = 0.
Therefore, the given equation is a quadratic equation.
ii.Here LHS x(x + 1) + 8 = x
2
+ x + 8 and RHS = (x + 2)(x – 2) = x
2
– 4
Therefore, x
2
+ x + 8 = x
2
– 4
x
2
+ x + 8 - x
2
+ 4 = 0
i.e.,x + 12 = 0
It is not in the form of ax
2
+ bx + c = 0.
Therefore, the given equation is not a quadratic equation.
iii.Here, LHS = x (2x + 3) = 2x
2
+ 3x
So, x (2x + 3) = x
2
+ 1 can be rewritten as
2x
2
+ 3x = x
2
+ 1
Therefore, we get x
2
+ 3x – 1 = 0
It is in the form of ax
2
+ bx + c = 0.
So, the given equation is a quadratic equation.
iv.Here, LHS = (x + 2)
3
= (x + 2)
2
(x + 2)
= (x
2
+ 4x + 4) (x + 2)
= x
3
+ 2x
2
+ 4x
2
+ 8x + 4x + 8
= x
3
+ 6x
2
+ 12x + 8
Therefore, (x + 2)
3
= x
3
– 4 can be rewritten as
x
3
+ 6x
2
+ 12x + 8 = x
3
– 4
i.e.,6x
2
+ 12x + 12 = 0 or, x
2
+ 2x + 2 = 0
It is in the form of ax
2
+ bx + c = 0.
So, the given equation is a quadratic equation.
Remark : Be careful! In (ii) above, the given equation appears to be a quadratic equation, but
it is not a quadratic equation.
In (iv) above, the given equation appears to be a cubic equation (an equation of degree
3) and not a quadratic equation. But it turns out to be a quadratic equation. As you can see, often
we need to simplify the given equation before deciding whether it is quadratic or not.

109. Mathematic1 0 6
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EXERCISE - 5.1
1.Check whether the following are quadratic equations :
i.(x + 1)
2
= 2(x – 3) ii.x
2
– 2x = (–2) (3 – x)
iii.(x – 2)(x + 1) = (x – 1)(x + 3) iv.(x – 3)(2x +1) = x(x + 5)
v.(2x – 1)(x – 3) = (x + 5)(x – 1)vi.x
2
+ 3x + 1 = (x – 2)
2
vii.(x + 2)
3
= 2x (x
2
– 1) viii.x
3
– 4x
2
– x + 1 = (x – 2)
3
2.Represent the following situations in the form of quadratic equations :
i.The area of a rectangular plot is 528 m
2
. The length of the plot (in metres) is one more
than twice its breadth. We need to find the length and breadth of the plot.
ii.The product of two consecutive positive integers is 306. We need to find the integers.
iii.Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years
from now will be 360. We need to find Rohan’s present age.
iv.A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h
less, then it would have taken 3 hours more to cover the same distance. We need to
find the speed of the train.
5.3 SOLUTION OF A QUADRATIC EQUATION BY FACTORISATION
We have learned to represent some of the daily life situations mathematically in the form of
quadratic equation with an unknown variable ‘x’.
Now we need to find the value of x.
Consider the quadratic equation 2x
2
– 3x + 1 = 0. If we replace x by 1. Then, we get
(2 × 1
2
) – (3 × 1) + 1 = 0 = RHS of the equation since 1 satisfied the equation . We say that 1
is a root of the quadratic equation 2x
2
– 3x + 1 = 0.
x = 1 is a solution of the quadratic equation.
This also means that 1 is a zeros of the quadratic polynomial 2x
2
– 3x + 1.
In general, a real number  is called a root of the quadratic equation ax
2
+ bx + c = 0,
a 0 if a
2
+ b + c = 0. We also say that x =  is a solution of the quadratic equation,
or  satisfies the quadratic equation.
Note that the zeroes of the quadratic polynomial ax
2
+ bx + c and the roots of the
quadratic equation ax
2
+ bx + c = 0 are the same.
We have observed, in Chapter 3, that a quadratic polynomial can have at most two zeroes.
So, any quadratic equation can have atmost two roots. (Why?)

110. Quadratic Equations1 0 7
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We have learnt in Class-IX, how to factorise quadratic polynomials by splitting their middle
terms. We shall use this knowledge for finding the roots of a quadratic equation. Let us see.
Example-3.Find the roots of the equation 2x
2
– 5x + 3 = 0, by factorisation.
Solution : Let us first split the middle term. Recall that if a quadratic equation is ax
2
+ bx + c =
0 then to split the middle term we have to find two numbers p and q such that p + q = b and
p × q = a × c so after compair both the equation we have a = 2, b = -5 & c = 3 and then p +
q = b = –5 and p × q = a × c = 2 × 3 = 6
For this we have to see the pairs of factor of 6 they are (1, 6), (–1, –6); (2, 3); (–2, –3);
from the above we can find that –2, –3 will satisfy p + q = –5 and p × q = 6.
The middle term ‘–5x’ can be written as ‘–2x – 3x’.
So, 2x
2
– 5x + 3 = 2x
2
– 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1)
Now, 2x
2
– 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0.
So, the values of x for which 2x
2
– 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0,
i.e., either 2x – 3 = 0 or x – 1 = 0. (zero - product property)
Now, 2x – 3 = 0 gives x =
3
2
and x – 1 = 0 gives x = 1.
So, x =
3
2
and x = 1 are the solutions of the equation.
In other words, 1 and
3
2
are the roots of the equation 2x
2
– 5x + 3 = 0.
TRY THIS
Verify that these are the roots of the given equation.
Note that we have found the roots of 2x
2
– 5x + 3 = 0 by factorising 2x
2
– 5x + 3
into two linear factors and equating each factor to zero.
Example 4 : Find the roots of the quadratic equation x-
1
3x
=
1
6
Solution : We have x-
1
3x
=
1
6
 6x
2
- x - 2 = 0

111. Mathematic1 0 8
Free Distribution by A.P. Government
6x
2
– x – 2 = 6x
2
+ 3x – 4x – 2
= 3x (2x + 1) – 2 (2x + 1)
= (3x – 2)(2x + 1)
The roots of 6x
2
– x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0
Therefore, 3x – 2 = 0 or 2x + 1 = 0,
i.e.,x =
2
3
or x =
1
2

Therefore, the roots of 6x
2
– x – 2 = 0 are
2
3
and
1
2
.
We verify the roots, by checking that
2
3
and
1
2
 satisfy 6x
2
– x – 2 = 0.
Example-5.Find the width of the space for spectators discussed in section 5.1.
Solution : In Section 5.1, we found that if the width of the space for spectators is x m., than x
satisfies the equation 2x
2
+ 45x - 47 = 0. Applying the factorisation method we write this equation
as:-
2x
2
- 2x + 47x - 47 = 0
2x (x - 1) + 47 (x - 1) = 0
i.e.,(x - 1) (2x + 47) = 0
So, the roots of the given equation are x = 1 or x =
47
2

. Since ‘x’ is the width of space
of the spectators it cannot be negative.
Thus, the width is 1 m.
EXERCISE - 5.2
1.Find the roots of the following quadratic equations by factorisation:
i.x
2
– 3x – 10 = 0ii.2x
2
+ x – 6 = 0 iii.
2
2 7 5 2 0  x x
iv.
2 1
2 0
8
  x x v.100x
2
– 20x + 1 = 0vi.x(x + 4) = 12
vii.3x
2
– 5x + 2 = 0viii.
3
2 x
x
ix.3(x – 4)
2
– 5(x – 4) = 12

112. Quadratic Equations1 0 9
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2.Find two numbers whose sum is 27 and product is 182.
3.Find two consecutive positive integers, sum of whose squares is 613.
4.The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the
other two sides.
Find the quadratic equation which have roots 3 and
1
2
5.A cottage industry produces a certain number of pottery articles in a day. It was observed
on a particular day that the cost of production of each article (in rupees) was 3 more than
twice the number of articles produced on that day. If the total cost of production on that
day was Rs 90, find the number of articles produced and the cost of each article.
6.Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40
square meters.
7.The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq.cm
then find its base and altitude.
8.Two trains leave the railway station at the same time. The first train travels towards west
and the second train towards north. The first train travels 5 km/hr faster than the second
train. If after two hours they are 50 km. apart find the average speed of each train.
9.In a class of 60 students, each boy contributed rupees equal to the number of girls and
each girl contributed rupees equal to the number of boys. If the total money then collected
was D1600. How many boys are there in the class?
10.A motor boat heads upstream a distance of 24km an a river whose current is running at 3
km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained
a constant speed, what was its speed?
5.4 SOLUTION OF A QUADRATIC EQUATION BY COMPLETING THE SQUARE
In the previous section, we have learnt method of factorisation forobtaining the roots of a
quadratic equation. Is method of factorization is applicable to all types of quadratic equation?
Let us try to solve x
2
+ 4x - 4 = 0 by factorisation
x
2
+ 4x - 4 = 0 cannot be solved by factorisation.
Since for no p and q :p + q = 4 and
p × q = -4
Therefore, we shall study another method.

113. Mathematic1 1 0
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Consider the following situation
The product of Sunita’s age (in years) two years ago and her age four years from now is
one more than twice her present age. What is her present age?
To answer this, let her present age (in years) be x years. Age before two year = x - 2 &
age after four year = x + 4 then the product of both the ages is (x – 2)(x + 4).
Therefore,(x – 2)(x + 4) = 2x + 1
i.e., x
2
+ 2x – 8 = 2x + 1
i.e., x
2
– 9 = 0
So, Sunita’s present age satisfies the quadratic equation x
2
– 9 = 0.
We can write this as x
2
= 9. Taking square roots, we get x = 3 or x = – 3. Since the age is
a positive number, x = 3.
So, Sunita’s present age is 3 years.
Now consider the quadratic equation (x + 2)
2
– 9 = 0. To solve it, we can write it as
(x + 2)
2
= 9. Taking square roots, we get x + 2 = 3 or x + 2 = – 3.
Therefore,x = 1 or x = –5
So, the roots of the equation (x + 2)
2
– 9 = 0 are 1 and – 5.
In both the examples above, the term containing x is completely inside a square, and we
found the roots easily by taking the square roots. But, what happens if we are asked to solve the
equation x
2
+ 4x – 4 = 0. And it cannot be solved by factorisation also.
So, we now introduce the method of completing the square. The idea behind this
method is to adjust the left side of the quadratic equation so that it becomes a perfect
square.
The process is as follows:
x
2
+ 4x – 4 = 0
 x
2
+ 4x = 4
x
2
+ 2. x . 2 = 14
Now, the LHS is in the form of a
2
+ 2ab. If we added b
2
it becomes as a
2
+ 2ab + b
2
which is perfect square. So, by adding b
2
= 2
2
= 4 to both sides we get,
x
2
+ 2.x.2 + 2
2
= 4 + 4
 (x + 2)
2
= 8 x + 2 = 8
 x = –2 2 2

114. Quadratic Equations1 1 1
Free Distribution by A.P. Government
Now consider the equation 3x
2
– 5x + 2 = 0. Note that the coefficient of x
2
is not 1 So we
divide the entire equation by 3 so that the coefficient of x
2
is 1

25 2
0
3 3
  x x

25 2
3 3

 x x

2 5 2
2. .
6 3

 x x

2 2
2 5 5 2 5
2. .
6 6 3 6
   
   
   
   
x x
2
3
add both side
6
 
 
  
 
 
 
2
5 2 25
6 3 36
 
  
 
 
x
  
2
12 2 25 15
=
6 36
x
   
 
 
2
5 24 25
=
6 36
x
  
 
 
2
5 1
6 36
 
 
 
 
x (take both side square root)
5 1
6 6
  x
So,
5 1
6 6
 x or
5 1
6 6
 x
Therefore, x = 1 or x =
4
6
i.e., x = 1 or x =
2
3
Therefore, the roots of the given equation are 1 and
2
3
.
From the above examples we can deduce the following algorithm for computing the square.
Algorithm : Let the quadratic equation by ax
2
+ bx + c = 0
Step-1 : Divide each side by ‘a’

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Step-2 : Rearrange the equation so that constant term c/a is on the right side. (RHS)
Step-3 : Add
2
1
2
  
  
  
b
a
to both sides to make LHS, a perfect square.
Step-4 : Write the LHS as a square and simplify the RHS.
Step-5 : Solve it.
Example-6.Find the roots of the equation 5x
2
– 6x – 2 = 0 by the method of computing the
square.
Solution : Given : 5x
2
– 6x – 2 = 0
Now we follow the Algoritm
Step-1 :
26 2
0
5 5
  x x (Dividing both sides by 5)
Step-2 :
26 2
5 5
 x x
Step-3 :
2 2
26 3 2 3
5 5 5 5
   
   
   
   
x x
2
3
Adding to both sides
5
 
 
   
 
 
 
Step-4 :
2
3 2 9
5 5 25
 
  
 
 
x
Step-5 :
2
3 19
5 25
 
 
 
 
x
3 19 19
5 25 5
    x
3 19
5 5
 x or
3 19
5 5
 x

3 19
5

x or
3 19
5

x
Example-7.Find the roots of 4x
2
+ 3x + 5 = 0 by the method of computing the square.

116. Quadratic Equations1 1 3
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Solution : Given 4x
2
+ 3x + 5 = 0

23 5
0
4 4
  x x

23 5
4 4

 x x

2 2
23 3 5 3
4 8 4 8
   
   
   
   
x x

2
3 5 9
8 4 64
 
  
 
 
x
2
3 71
0
8 64
 
  
 
 
x
But
2
3
8
 

 
 
x cannot be negative for any real value of x (Why?). So, there is no real value
of x satisfying the given equation. Therefore, the given equation has no real roots.
TRY THESE
Solve the equations by computing the square
(i)x
2
 10x + 9 = 0 (ii)x
2
 5x + 5= 0
(iii)x
2
+ 7x  6 = 0
We have solved several examples with the use of the method of ‘computing the square.’
Now, let us apply this method in standard form of quadratic equation ax
2
+bx+c=0 (a0).
Dividing the equation throughout by a,
we get
2
0  
b c
x x
a a
This is the same as
2 2
0
2 2
   
   
   
   
b b c
x
a a a
2
adding & subtract
2
b
a
 
 
   
   
2
2
2
0
2 4
b b c
x
a a a
 
   
 
 

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i.e.,
2 2
2
4
0
2 4
 
  
 
 
b b ac
x
a a
So, the roots of the given equation are the same as those of
2 2
2
4
0
2 4
 
  
 
 
b b ac
x
a a
i.e., those of
2 2
2
4
2 4
 
 
 
 
b b ac
x
a a
(1)
If b
2
 4ac > 0, then by taking the square roots in (1), we get
2
4
2 2
 
 
b b ac
x
a a
Therefore,
2
4
2
 

b b ac
x
a
So, the roots of ax
2
+ bx + c = 0 are
2
4
2
  b b ac
a
and
2
4
2
  b b ac
a
,
if b
2
– 4ac > 0.
If b
2
– 4ac < 0, the equation will have no real roots. (Why?)
Thus, if b
2
– 4ac > 0, then the roots of the quadratic equation ax
2
+ bx + c = 0 are
given by
2
4
2
b b ac
a
  
.
This formula for finding the roots of a quadratic equation is known as the quadratic formula.
Let us consider some examples by using quadratic formula.
Example-8.Solve Q. 2(i) of Exercise 4.1 by using the quadratic formula.
Solution : Let the breadth of the plot be x metres.
Then the length is (2x + 1) metres.
Since area of rectangular plot is 528 m
2
Can write x(2x + 1) = 528, i.e., 2x
2
+ x – 528 = 0.
This is in the form of ax
2
+ bx + c = 0, where a = 2, b = 1, c = – 528.
So, the quadratic formula gives us the solution as

118. Quadratic Equations1 1 5
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1 1 4(2)(528) 1 4225 1 65
4 4 4
      
  x
i.e.,
64
4
x or
66
4

x
i.e.,x = 16 or x =
33
2

Since x cannot be negative. So, the breadth of the plot is 16 metres and hence, the length
of the plot is (2x + 1) = 33m.
You should verify that these values satisfy the conditions of the problem.
THINK & WRITE
We have two methods to solve a quadratic equation. Among these two, which method
would you like to use do you? Why?
Example-9.Find two consecutive odd positive integers, sum of whose squares is 290.
Solution : Let first consecutive odd positive integers be x. Then, the second integer will be x +
2. According to the question,
x
2
+ (x + 2)
2
= 290
i.e.,x
2
+ x
2
+ 4x + 4 = 290
i.e.,2x
2
+ 4x – 286 = 0
i.e.,x
2
+ 2x – 143 = 0
which is a quadratic equation in x.
Using the quadratic formula
2
4
=
2
b b ac
x
  
we get,
2 4 572 2 576 2 24
2 2 2
      
  x
i.e.,x = 11 or x = – 13
But x is given to be an odd positive integer. Therefore, x – 13, x = 11.
Thus, the two consecutive odd integers are 11 and (x + 2) = 11 + 2 = 13.
Check : 11
2
+ 13
2
= 121 + 169 = 290.

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Example-10. A rectangular park is to be designed whose breadth is 3 m less than its length. Its
area is to be 4 square metres more than the area of a park that has already been made in the
shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude
12 m (see Fig. 5.3). Find its length and breadth.
Solution : Let the breadth of the rectangular park be x m.
So, its length = (x + 3) m.
Therefore, the area of the rectangular park = x(x + 3) m
2
= (x
2
+ 3x) m
2
.
Now, base of the isosceles triangle = x m.
Therefore, its area =
1
2
× x × 12 = 6 x m
2
.
According to our requirements,
x
2
+ 3x = 6x + 4
i.e., x
2
– 3x – 4 = 0
Using the quadratic formula, we get
3 25 3 5
2 2
 
 x = = 4 or – 1
But x – 1 (Why?). Therefore, x = 4.
So, the breadth of the park = 4m and its length will be x + 3 = 4 + 3 = 7m.
Verification : Area of rectangular park = 28 m
2
,
area of triangular park = 24 m
2
= (28 – 4) m
2
Example-11. Find the roots of the following quadratic equations, if they exist, using
the quadratic formula:
(i) x
2
+ 4x + 5 = 0 (ii) 2x
2
– 22x + 1 = 0
Solution :
(i)x
2
+ 4x + 5 = 0. Here, a = 1, b = 4, c = 5. So, b
2
– 4ac = 16 – 20 = – 4 < 0.
Since the square of a real number cannot be negative, therefore
2
4b ac will not
have any real value.
So, there are no real roots for the given equation.
(ii)2x
2
– 22x + 1 = 0. Here, a = 2, b = 22, c = 1.
12
x
+
3
x

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Free Distribution by A.P. Government
So, b
2
– 4ac = 8 – 8 = 0
Therefore,
2 2 0 2
0
4 2

  x i.e., x =
1
2
.
So, the roots are
1 1
, .
2 2
Example-12. Find the roots of the following equations:
(i)
1
3, 0  x x
x
(ii)
1 1
3, 0,2
2
  

x
x x
Solution :
(i)
1
3 x
x
. Multiplying whole eg. by x, we get
x
2
+ 1 = 3x
i.e., x
2
– 3x + 1 = 0, which is a quadratic equation.
Here, a = 1, b = – 3, c = 1
So, b
2
– 4ac = 9 – 4 = 5 > 0
Therefore,
3 5
2

x (why ?)
So, the roots are
3 5
2

and
3 5
.
2

(ii)
1 1
3, 0, 2.
2
  

x
x x
As x  0, 2, multiplying the equation by x (x – 2), we get
(x – 2) – x = 3x (x – 2)
= 3x
2
– 6x
So, the given equation reduces to 3x
2
– 6x + 2 = 0, which is a quadratic equation.
Here,a = 3, b = – 6, c = 2. So, b
2
– 4ac = 36 – 24 = 12 > 0
Therefore,
6 12 6 2 3 3 3
.
6 6 3
  
  x

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So, the roots are
3 3
3

and
3 3
.
3

Example-13. A motor boat whose speed is 18 km/h. Still water takes 1 hour more to go 24 km
upstream than to return downstream to the same spot. Find the speed of the stream.
Solution : Let the speed of the stream be x km/h.
Therefore, the speed of the boat upstream = (18 – x) km/h and the speed of the boat
downstream = (18 + x) km/h.
The time taken to go upstream =
distance
speed
=
24
18x
hours.
Similarly, the time taken to go downstream =
24
18x
hours.
According to the question,
24 24
1
18 18
 
 x x
i.e., 24(18 + x) – 24(18 – x) = (18 – x) (18 + x)
i.e., x
2
+ 48x – 324 = 0
Using the quadratic formula, we get
2
48 48 1296 48 3600
2 2
    
 x

48 60
6
2
 
  or 54
Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = – 54.
Therefore, x = 6 gives the speed of the stream as 6 km/h.
EXERCISE - 5.3
1.Find the roots of the following quadratic equations, if they exist, by the method of computing
the square:
i.2x
2
+ x – 4 = 0 ii.
2
4 4 3 3 0  x x
iii.5x
2
 7x  6 = 0 iv.x
2
+ 5 = 6x

122. Quadratic Equations1 1 9
Free Distribution by A.P. Government
2.Find the roots of the quadratic equations given in Q.1 above by applying the quadratic
formula.
3.Find the roots of the following equations:
(i)
1
3, 0  x x
x
(ii)
1 1 11
, 4,7
4 7 30
   
 
x
x x
4.The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now
is
1
3
. Find his present age.
5.In a class test, the sum of Shefali’s marks in Mathematics and English is 30. If she got 2
marks more in Mathematics and 3 marks less in English, the product of her marks would
have been 210. Find her marks in the two subjects.
6.The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer
side is 30 metres more than the shorter side, find the sides of the field.
7.The difference of squares of two numbers is 180. The square of the smaller number is 8
times the larger number. Find the two numbers.
8.A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would
have taken 1 hour less for the same journey. Find the speed of the train.
9.Two water taps together can fill a tank in
3
9
8
hours. The tap of larger diameter takes 10
hours less than the smaller one to fill the tank separately. Find the time in which each tap
can separately fill the tank.
10.An express train takes 1 hour less than a passenger train to travel 132 km between Mysore
and Bangalore (without taking into consideration the time they stop at intermediate stations).
If the average speed of the express train is 11km/h more than that of the passenger train,
find the average speed of the two trains.
11.Sum of the areas of two squares is 468 m
2
. If the difference of their perimeters is 24 m, find
the sides of the two squares.
12.A ball is thrown vertically upward from the top of a building 96 feet tall with an initial
velocity 80 m/second. The distance ‘s’ of the ball from the ground after t seconds is S =
96 + 80t  16t
2
. After how may seconds does the ball strike the ground.
13.If a polygon of ‘n’ sides has
1
2
n (n3) diagonals. How many sides will a polygon with 65
diagonals have ? Is there a polygon with 50 diagonals?

123. Mathematic1 2 0
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5.5 NATURE OF ROOTS
In the previous section, we have seen that the roots of the equation ax
2
+ bx + c = 0 are
given by
2
4
2
  

b b ac
x
a
Now let us try to understand the nature of roots.
Case-1 :If b
2
- 4ac > 0;
We get tood distinct real roots
2
4
2
b b ac
a
  
,
2
4
2
b b ac
a
  
going back to the
polynomials, zeros are those points where value of polynomial becomes zero or we can say that
the curve of quadratic polynomial cuts the X-axis.
Similarly, roots of a quadratic equation are those points where the curve cuts the X-axis.
Case-2 :If b
2
- 4ac = 0
x =
0
2
b
a
 
So,x =
2
b
a

,
2
b
a

Figure shows curve of the quadratic equation touching X-axis at one point.
Case-3 :b
2
- 4ac < 0
There are no real roots. Roots are imaginary.

124. Quadratic Equations1 2 1
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In this case graph does not touch the X-axis at all. So, no real roots are these.
Since b
2
– 4ac determines whether the quadratic equation ax
2
+ bx + c = 0 has real roots
or not, b
2
– 4ac is called the discriminant of this quadratic equation.
So, a quadratic equation ax
2
+ bx + c = 0 has
i.two distinct real roots, if b
2
– 4ac > 0,
ii.two equal real roots, if b
2
– 4ac = 0,
iii.no real roots, if b
2
– 4ac < 0.
Let us consider some examples.
Example-14. Find the discriminant of the quadratic equation 2x
2
– 4x + 3 = 0, and hence find
the nature of its roots.
Solution : The given equation is in the form of ax
2
+ bx + c = 0, where a = 2, b = – 4 and
c = 3. Therefore, the discriminant
b
2
– 4ac = (– 4)
2
– (4 × 2 × 3) = 16 – 24 = – 8 < 0
So, the given equation has no real roots.
Example-15. A pole has to be erected at a point on the boundary of a circular park of diameter
13 metres in such a way that the differences of its distances from two diametrically opposite fixed
gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from
the two gates should the pole be erected?
Solution : Let us first draw the diagram.
Let P be the required location of the pole. Let the distance of the
pole from the gate B be x m, i.e., BP = x m. Now the difference of the
distances of the pole from the two gates = AP – BP (or, BP – AP)= 7 m.
Therefore, AP = (x + 7) m.
Now, AB = 13m, and since AB is a diameter,
APB = 90
0
(Why?)
Therefore,AP
2
+ PB
2
= AB
2
(By Pythagoras theorem)
i.e., (x + 7)
2
+ x
2
= 13
2
i.e., x
2
+ 14x + 49 + x
2
= 169
i.e.,2x
2
+ 14x – 120 = 0
A
B
P
13

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So, the distance ‘x’ of the pole from gate B satisfies the equation
x
2
+ 7x – 60 = 0
So, it would be possible to place the pole if this equation has real roots. To see if this is so or not,
let us consider its discriminant. The discriminant is
b
2
– 4ac = 7
2
– 4 × 1 × (– 60) = 289 > 0.
So, the given quadratic equation has two real roots, and it is possible to erect the pole on
the boundary of the park.
Solving the quadratic equation x
2
+ 7x – 60 = 0, by the quadratic formula, we get
7 289 7 17
2 2
   
 x
Therefore, x = 5 or – 12.
Since x is the distance between the pole and the gate B, it must be positive.
Therefore, x = – 12 will have to be ignored. So, x = 5.
Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the
gate B and 12m from the gate A.
TRY THESE
1.Explain the benefits of evaluating the discriminant of a quadritic equation before attempting
to solve it. What does its value signifies?
2.Make up true quadratic equations one having two distinct real solutions, one having no real
solution and one having exactly one real solution.
Example-16. Find the discriminant of the equation
2 1
3 2
3
 x x = 0 and hence find the nature
of its roots. Find them, if they are real.
Solution : Here a = 3, b = – 2 and c =
1
3
Therefore, discriminant
2 2 1
4 ( 2) 4 3 4 4 0.
3
        b ac
Hence, the given quadratic equation has two equal real roots.
The roots are , ,
2 2
 b b
a a
i.e.,
2 2
, ,
6 6
i.e.,
1 1
, .
3 3

126. Quadratic Equations1 2 3
Free Distribution by A.P. Government
EXERCISE - 5.4
1.Find the nature of the roots of the following quadratic equations. If the real roots exist,
find them:
(i)2x
2
– 3x + 5 = 0 (ii)
2
3 4 3 4 0  x x
(iii)2x
2
– 6x + 3 = 0
2.Find the values of k for each of the following quadratic equations, so that they have two
equal roots.
(i)2x
2
+ kx + 3 = 0 (ii)kx (x – 2) + 6 = 0
3.Is it possible to design a rectangular mango grove whose length is twice its breadth, and the
area is 800 m
2
? If so, find its length and breadth.
4.The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in
years was 48. Is the following situation possible? If so, determine their present ages.
5.Is it possible to design a rectangular park of perimeter 80 m. and area 400 m
2
? If so, find
its length and breadth.
OPTIONAL EXERCISE
[This exercise is not meant for examination]
1.Some points are plotted on a plane. Each point joined with all remaining points by line
segments. Find the number of points if the number of line segments are 10.
2.A two digit number is such that the product of the digits is 8. When 18 is added to the
number they interchange their places. Determine the number.
3.A piece of wire 8 m. in length. It cut into two pieces, the each piece will be bent into a
square. Where should the cut in the wire be made if the sum of the areas of these squares
is to be 2 m
2
?
2 2 2 2
8
8, 2 2
4 4
 
       
       
      
        
 
x y x x
x y
y x
.
5.Vinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by
himself can complete the job in 5 days less than Praveen. How long will it take vinay to
complete the job by himself.
6.Show that the sum of roots of a quadratic equation is ‘
b
a
’.

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7.Show that the product of the roots of a quadratic equation is ‘
c
a
’.
8.The denominator of a fraction is one more than twice the numerator. If the sum of the
fraction and its reciprocal is 2
16
21
, find the fraction.
WHAT WE HAVE DISCUSSED
In this chapter, you have studied the following points:
1.Standard form of quadratic equation in variable x is ax
2
+ bx + c = 0, where a, b, c are
real numbers and a  0.
2.A real number  is said to be a root of the quadratic equation ax
2
+ bx + c = 0, if
a
2
+ + b c = 0. The zeroes of the quadratic polynomial ax
2
+ bx + c and the roots of
the quadratic equation ax
2
+ bx + c = 0 are the same.
3.If we can factorise ax
2
+ bx + c, a 0, into a product of two linear factors, then the roots
of the quadratic equation ax
2
+ bx + c = 0 can be found by equating each factor to zero.
4.A quadratic equation can also be solved by the method of completing the square.
5.Quadratic formula: The roots of a quadratic equation ax
2
+ bx + c = 0 are given by
2
4
,
2
  b b ac
a
provided b
2
– 4ac > 0.
6.A quadratic equation ax
2
+ bx + c = 0 has
(i)two distinct real roots, if b
2
– 4ac > 0,
(ii)two equal roots (i.e., coincident roots), if b
2
– 4ac = 0, and
(iii)no real roots, if b
2
– 4ac < 0.

128. 6.1 INTRODUCTION
You must have observed that in nature, many things follow a certain pattern such as
the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirds on
a pineapple and on a pine cone etc.
Can you see a pattern in each of the given example? We can see the natural patterns
have a repetition which is not progressive. The identical petals of the sunflower are
equidistantly grown. In a honeycomb identical hexagonal shaped holes are arranged
symmetrically around each hexagon. Similarly, you can find out other natural patterns in
spirds of pineapple....
You can look for some other patterns which occur in our day-to-day life. Some
examples are:
(i)List of the last digits (digits in unit place) taken from the values of 4, 4
2
, 4
3
, 4
4
, 4
5
, 4
6
.....
is
4, 6, 4, 6, 4, 6, ......
(ii)Mary is doing problems on patterns as part of preparing for a bank exam. One of them
is “find the next two terms in the following pattern”.
1, 2, 4, 8, 10, 20, 22 .......
(iii)Reena applied for a job and got selected. She has been offered a job with a starting
monthly salary of Rs. 8000, with an annual increment of Rs. 500 in her salary. Her
salary (in rupees) for to 1
st
, 2
nd
, 3
rd
... years will be 8000, 8500, 9000 ..... respectively.
(iv)The lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top.
The bottom rung is 45 cm in length. The lengths (in cm) of the 1
st
, 2
nd
, 3
rd
, .... 8
th
rung
from the bottom to the top are 45, 43, 41, 39, 37, 35, 33, 31 respectively.
Can you see any relationship between the terms in the pattern of numbers written
above?
Pattern given in example (i) has a relation of two numbers one after the other i.e. 4
and 6 are repeating alternatively.
Progressions6

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Now try to find out pattern in examples (ii) On example (iii) and (iv), the relationship
between the numbers in each list is constantly progressive. In the given list 8000, 8500, 9000,
.... each succeeding term is obtained by adding 500 to the preceding term.
Where as in 45, 43, 41, ..... each succeeding term is obtained by adding ‘-2’ to each
preceding term. New we can see some more examples of progressive patterns.
(a)In a savings seheme, the amount becomes
5
4
times of itself after 3 years.
The maturity amount (in Rupees) of an investment of Rs. 8000 after 3, 6, 9 and 12 years
will be respectively. 10000, 12500, 15625, 19531.25.
(b)The number of unit squares in squares with side 1, 2, 3, .... units are respectively.
1
2
, 2
2
, 3
2
, ....
(c)Rajitha put Rs. 1000 into her daughter’s money box when she was one year old and
increased the amount by Rs. 500 every year. The amount of money (in Rs.) in the box
on her 1
st
, 2
nd
, 3
rd
, 4
th
birthday would be.
1000, 1500, 2000, 2500, ..... respectively.
(d)The fraction of first, second, third ..... shaded regions of the squares in the following
figure will be respectively.
1 1 1 1
, , , , ....
4 16 64 256

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(e)A pair of rabbits are too young to produce in their first month. In the second, and
every subsequent month, they produce a new pair. Each new pair of rabbits produce
a new pair in their second month and in every subsequent month (see the figure
below). Assuming no rabbit dies, the number of pairs of rabbits at the start of the 1
st
,
2
nd
, 3
rd
, ....., 6
th
month, respectively are :
1, 1, 2, 3, 5, 8
In the examples above, we observe some patterns. In some of them, we find that the
succeeding terms are obtained by adding a fixed number or in other by multiplying with a
fixed number or in another, we find that they are squares of consecutive numbers and so on.
In this chapter, we shall discuss one of these patterns in which succeeding terms are
obtained by adding a fixed number to the preceding terms. We shall also see how to find their
n
th
term and the sum of n consecutive terms, and use this knowledge in solving some daily
life problems.
History : Evidence is found that Babylonians some 400 years ago, knew of Arithmetic and
geometric progressions. According to Boethins (570 AD), these progressions were known to
early Greek writers. Among the Indian mathematicians, Aryabhata (470 AD) was the first to
give formula for the sum of squares and cubes of natural number in his famous work.
Aryabhatiyam written around 499 A.D. He also gave the formula for finding the sum of n
terms of an Arithmetic Progression starting with p
th
term. Indian mathematician Brahmagupta
(598 AD), Mahavira (850 AD) and Bhaskala (1114-1185 AD) also considered the sums of
squares and cubes.
1
1
2
3
5
8

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6.2 ARITHMETIC PROGRESSIO NS
Consider the following lists of numbers :
(i)1, 2, 3, 4, . . . (ii)100, 70, 40, 10, . . .
(iii)– 3, –2, –1, 0, . . . (iv)3, 3, 3, 3, . . .
(v)–1.0, –1.5, –2.0, –2.5, . . .
Each of the numbers in the list is called a term.
Given a term, can you write the next term in each of the lists above? If so, how will you
write it? Perhaps by following a pattern or rule, let us observe and write the rule.
In (i), each term is 1 more than the term preceding it.
In (ii), each term is 30 less than the term preceding it.
In (iii), each term is obtained by adding 1 to the term preceding it.
In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding (or subtracting) 0 to the
term preceding it.
In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the term preceding it.
In all the lists above, we can observe that successive terms are obtained by adding or
substracting a fixed number to the preceding terms. Such list of numbers is said to form an
Arithmetic Progression ( AP ).
TRY THESE
(i)Which of these are Arithmetic Progression’s and why?
(a)2, 3, 5, 7, 8, 10, 15, ......(b)A, C, E, G, I, ......
(c)-1, -3, -5, -7, ......
(ii)Write 3 more Arithmetic Progression’s.
6.2.1 WHAT IS AN ARITHM ETIC PROGRESSION?
We observe that an arithmetic progression is a list of numbers in which each term
is obtained by adding a fixed number to the preceding term except the first term.
This fixed number is called the common difference of the AP. Remember that it can be
positive, negative or zero.
Let us denote the first term of an AP by a
1
, second term by a
2
, . . ., nth term by a
n
and the
common difference by d. Then the AP becomes a
1
, a
2
, a
3
, . . ., a
n
.
So, a
2
– a
1
= a
3
– a
2
= . . . = a
n
– a
n – 1
= d.

132. Progressions1 2 9
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Let us see some more examples of AP :
(a)Heights ( in cm ) of some students of a school standing in a queue in the morning assembly
are 147 , 148, 149, . . ., 157.
(b)Minimum temperatures ( in degree celsius ) recorded for a week, in the month of January
in a city, arranged in ascending order are
– 3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5
(c)The balance money ( in Rs ) after paying 5% of the total loan of Rs 1000 every month is
950, 900, 850, 800, . . ., 50.
(d)Cash prizes ( in Rs ) given by a school to the toppers of Classes I to XII are 200, 250,
300, 350, . . ., 750 respectively.
(e)Total savings (in Rs) after every month for 10 months when Rs 50 are saved each month
are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
TRY THESE
1. Think how each of the list given above form an A.P. Discuss with your friends.
2. Find the common difference in each of thire in which is the common difference positive.
3. Make 2 AP’s each, where common difference is positive (in one small and in one large)
and one where is negative.
General form of AP : Can you see that all AP’s can be written as.
a, a + d, a + 2d, a + 3d, . . .
This general form can represent any arithmetic progression. In this a is the first term and d
is the common difference.
For example in 1, 2, 3, 4, 5, ....
The first terms is 1 and the common difference is also 1 in 2, 4, 6, 8, 10, ...... What is the
first term and common difference?
TRY THIS
(i)Make 3 more AP’s and write its first term and the common difference.
(ii)Make the following figures with match sticks

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(iii)Write down the number of match sticks required for each figure.
(iv)Does the list of these numbers form an A.P.?
(v)Can you find a common difference in members of the list?
6.2.2 PARAM ETERS OF A ARITHMETIC PROGRESSIONS
Note that in examples (a) to (e) above, there are only a finite number of terms. Such an AP
is called a finite AP. Also note that each of these Arithmetic Progressions (APs) has a last term.
The APs in examples (i) to (v) in this section, are not finite APs and so they are called infinite
Arithmetic Progressions. Such APs are never ending and do not have a last term.
TRY THESE
Write three examples finite and infinite A.P. separately.
Now, to know about an AP, what is the minimum information that you need? Is it enough
to know the first term? Or, is it enough to know only the common difference?
We can see that we will need to know both – the first term a and the common difference
d. These two parameters are sufficient for us to complete the Arithmetic Progression.
For instance, if the first term a is 6 and the common difference d is 3, then the AP is
6, 9,12, 15, . . .
and if a is 6 and d is – 3, then the AP is
6, 3, 0, –3, . . .
Similarly, when
a = – 7,d = – 2, the AP is – 7, – 9, – 11, – 13, . . .
a = 1.0,d = 0.1, the AP is 1.0, 1.1, 1.2, 1.3, . . .
a = 0,d = 1
1
2
, the AP is 0, 1
1
2
, 3, 4
1
2
, 6, . . .
a = 2,d = 0, the AP is 2, 2, 2, 2, . . .
So, if you know what a and d are, you can list the AP.
Let us try other way. If you are given a list of numbers, how you say whether it is an A.P.
or not?
For example, for any list of numbers :
6, 9, 12, 15, . . . ,

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We check the difference of the succeeding terms. In the given list we have a
2
– a
1
= 9 – 6 = 3,
a
3
– a
2
= 12 – 9 = 3,
a
4
– a
3
= 15 – 12 = 3
We see that
2 1 3 2 1 3
... 3a a a a a a     
Here the difference of any two consecutive terms in each case is 3. So, the given list is an
AP whose first term a is 6 and common difference d is 3.
For the list of numbers : 6, 3, 0, – 3, . . .,
a
2
– a
1
= 3 – 6 = – 3,
a
3
– a
2
= 0 – 3 = – 3
a4 – a
3
= –3 – 0 = –3
2 1 3 2 4 3
3a a a a a a     
Similarly, this is also an AP whose first term is 6 and the common difference is –3.
So, we see that if the difference between succeeding terms is constant Arithmetic Progression.
In general, for an AP a
1
, a
2
, . . ., a
n
, we can say
d = a
k + 1
– a
k
where a
k + 1
and a
k
are the (k + 1)th and the kth terms respectively.
Consider the list of numbers 1, 1, 2, 3, 5, . . . . By looking at it, you can tell that the
difference between any two consecutive terms is not the same. So, this is not an AP.
Note : To find d in the AP : 6, 3, 0, – 3, . . ., we have subtracted 6 from 3 and not 3 from 6.
We have to subtract the k
th
term from the (k + 1) th term irrespective of which is smaller. We also
know that if we have an AP in which finding the difference between any two consequetive terms
is enough, we do not have to check the difference between all terms one by one.
DO THIS
1.Take any Arithmetic Progression.
2.Add a fixed number to each and every term of A.P. Write the resulting numbers as a list.
3.Similarly subtract a fixed number from each and every term of AP. Write the resulting
numbers as a list.
4.Multiply and divide each term of AP by a fixed number and write the resulting numbers as
a list.
5.Check whether the resulting lists are AP in each case.
6.What is your conclusion?

135. Mathematic1 3 2
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Finding checking lists for AP and finding common difference. We have seen what an AP is
and how do we check a serices for be A.P. we look at after examples.
Example-1. For the AP :
1 1 3 5
, , ,
4 4 4 4
  
........, write the first term a and the common
difference d. What would be the seventh term?
Solution : Here, a =
1
4
; d =
1 1 1
4 4 2
 
 
Remember that we can find d using any two consecutive terms, once we know that the
numbers are in AP.
The seventh term would be
5 1 1 1 11
4 2 2 2 4
 
   
Example-2. Which of the following form an AP?
(i)4, 10, 16, 22, . . . (ii) 1, – 1, – 3, – 5, . . .(iii) – 2, 2, – 2, 2, – 2, . . .
(iv)1, 1, 1, 2, 2, 2, 3, 3, 3, . . . (v) x, 2x, 3x, 4x ......
Solution : (i) We have a
2
– a
1
= 10 – 4 = 6
a
3
– a
2
= 16 – 10 = 6
a
4
– a
3
= 22 – 16 = 6
i.e., a
k + 1
– a
k
is the same every time.
So, the given list of numbers forms an AP with the common difference d = 6.
The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.
(ii)a
2
– a
1
= – 1 – 1 = – 2
a
3
– a
2
= – 3 – ( –1 ) = – 3 + 1 = – 2
a
4
– a
3
= – 5 – ( –3 ) = – 5 + 3 = – 2
i.e., a
k + 1
– a
k
is the same every time.
So, the given list of numbers forms an AP with the common difference d = – 2.
The next two terms are:
– 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9
(iii)a
2
– a
1
= 2 – (– 2) = 2 + 2 = 4
a
3
– a
2
= – 2 – 2 = – 4

136. Progressions1 3 3
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As a
2
– a
1
 a
3
– a
2
, the given list of numbers does not form an AP..
(iv)a
2
– a
1
= 1 – 1 = 0
a
3
– a
2
= 1 – 1 = 0
a
4
– a
3
= 2 – 1 = 1
Here, a
2
– a
1
= a
3
– a
2
 a
4
– a
3
.
So, the given list of numbers does not form an AP.
(v)We havea
2
– a
1
= 2x – x = x
a
3
– a
2
= 3x – 2x = x
a
4
– a
3
= 4x – 3x = x
i.e., a
k+1
– a
k
is same every time where K  {N} (Natural numbers).
So, the given list form an AP.
EXERCISE - 6.1
1. In which of the following situations, does the list of numbers involved make an arithmetic
progression, and why?
(i)The taxi fare after each km when the fare is ` 20 for the first km and rises by
` 8 for each additional km.
(ii)The amount of air present in a cylinder when a vacuum pump removes
1
4
of the
air remaining in the cylinder at a time.
(iii)The cost of digging a well, after every metre of digging, when it costs ` 150 for
the first metre and rises by ` 50 for each subsequent metre.
(iv)The amount of money in the account every year, when Rs 10000 is deposited at
compound interest at 8 % per annum.
2. Write first four terms of the AP, when the first term a and the common difference d are
given as follows:
(i)a = 10, d = 10 (ii)a = –2, d = 0
(iii)a = 4, d = – 3 (iv)a = – 1, d =
1
2
(v)a = – 1.25, d = – 0.25

137. Mathematic1 3 4
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3.For the following APs, write the first term and the common difference:
(i)3, 1, – 1, – 3, . . . (ii)– 5, – 1, 3, 7, . . .
(iii)
1 5 9 13
, , , ,....
3 3 3 3
(iv)0.6, 1.7, 2.8, 3.9, . . .
4. Which of the following are APs ? If they form an AP, find the common difference d and
write three more terms.
(i)2, 4, 8, 16, . . . (ii)
5 7
2, ,3, ,....
2 2
(iii)– 1.2, – 3.2, – 5.2, – 7.2, . . .(iv)– 10, – 6, – 2, 2, . . .
(v)3, 3 2,3 2 2,3 3 2,....   (vi)0.2, 0.22, 0.222, 0.2222, . . .
(vii)0, – 4, – 8, –12, . . . (viii)
1 1 1 1
, , , ,....
2 2 2 2
   
(ix)1, 3, 9, 27, . . . (x)a, 2a, 3a, 4a, . . .
(xi)a, a
2
, a
3
, a
4
, . . . (xii)2, 8, 18, 32,.....
(xiii)3, 6, 9, 12,.....
6.3 n
th
TERM OF AN ARITHMETIC PROGRESSION
Let us consider the offer to Reena who applied for a job and got selected. She has been
offered a starting monthly salary of ` 8000, with an annual increment of ` 500. What would be
her monthly salary of the fifth year?
To answer this, let us first see what her monthly salary for the second year would be.
It would be ` (8000 + 500) = ` 8500.
In the same way, we can find the monthly salary for the 3
rd
, 4
th
and 5
th
year by adding `
500 to the salary of the previous year.
So, the salary for the 3
rd
year = Rs (8500 + 500)
= Rs (8000 + 500 + 500)
= Rs (8000 + 2 × 500)
= Rs [8000 + (3 – 1) × 500](for the 3
rd
year)
= Rs 9000
Salary for the 4
th
year = Rs (9000 + 500)
= Rs (8000 + 500 + 500 + 500)

138. Progressions1 3 5
Free Distribution by A.P. Government
= Rs (8000 + 3 × 500)
= Rs [8000 + (4 – 1) × 500](for the 4
th
year)
= Rs 9500
Salary for the 5
th
year= Rs (9500 + 500)
= Rs (8000+500+500+500 + 500)
= Rs (8000 + 4 × 500)
= Rs [8000 + (5 – 1) × 500](for the 5
th
year)
= Rs 10000
Observe that we are getting a list of numbers
8000, 8500, 9000, 9500, 10000, . . .
These numbers are in Arithmetic Progression.
Looking at the pattern above, can we find her monthly salary in the 6
th
year? The 15
th
year?
And, assuming that she will still be working in the same job, what would the monthly salary be in
the 25
th
year? The salary would be by adding Rs 500 each time to the salary of the previous
year. Let us see its we can shorter the process. You may have already got some idea from the
way we have obtained the salaries above.
Salary for the 15
th
year= Salary for the 14
th
year + Rs 500
=
13 times
Rs 8000 500 500 500 ... 500 Rs.500
 
      
 
 

= Rs [8000 + 14 × 500]
= Rs [8000 + (15 – 1) × 500] = Rs 15000
i.e., First salary + (15 – 1) × Annual increment.
In the same way, her monthly salary for the 25
th
year would be
Rs [8000 + (25 – 1) × 500] = Rs 20000
= First salary + (25 – 1) × Annual increment
This example has given us an idea about how to write the 15
th
term, or the 25
th
term. We are
now ready to write the n
th
term.
Let a
1
, a
2
, a
3
, . . . be an AP whose first term a
1
is a and the common difference is d.
Then,
the second term a
2
= a + d = a + (2 – 1) d

139. Mathematic1 3 6
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the third term a
3
= a
2
+ d = (a + d) + d = a + 2d = a + (3 – 1) d
the fourth term a
4
= a3 + d = (a + 2d) + d = a + 3d = a + (4 – 1) d
. . . . . . . .
. . . . . . . .
Looking at the pattern, we can say that the nth term a
n
= a + (n – 1) d.
So, the n
th
term an of the AP with first term a and common difference d is given by
a
n
= a + (n – 1) d.
a
n
is also called the general term of the AP.
If there are m terms in the AP, then a
m
represents the last term which is sometimes also
denoted by l.
Finding terms of an AP : Using the above one we can find different terms of an arithemetic
progression.
Let us consider some examples.
Example-3. Find the 10th term of the AP : 5, 1, –3, –7 . . .
Solution : Here, a = 5, d = 1 – 5 = – 4 and n = 10.
We have a
n
= a + (n – 1) d
So, a
10
= 5 + (10 – 1) (–4) = 5 – 36 = –31
Therefore, the 10th term of the given AP is – 31.
Example-4. Which term of the AP : 21, 18, 15, . . . is – 81?
Is there any term 0? Give reason for your answer.
Solution : Here, a = 21, d = 18 – 21 = – 3 and if a
n
= – 81, we have to find n.
As a
n
= a + ( n – 1) d,
we have – 81 = 21 + (n – 1)(– 3)
– 81 = 24 – 3n
– 105 = – 3n
So, n = 35
Therefore, the 35th term of the given AP is – 81.
Next, we want to know if there is any n for which a
n
= 0. If such an n is there, then
21 + (n – 1) (–3) = 0,
i.e., 3(n – 1) = 21
i.e., n = 8

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So, the eighth term is 0.
Example-5. Determine the AP whose 3
rd
term is 5

and the 7
th
term is 9.
Solution : We have
a
3
= a + (3 – 1) d = a + 2d = 5 (1)
and a
7
= a + (7 – 1) d = a + 6d = 9 (2)
Solving the pair of linear equations (1) and (2), we get
a = 3, d = 1
Hence, the required AP is 3, 4, 5, 6, 7, . . .
Example-6. Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .
Solution : We have :
a
2
– a
1
= 11 – 5 = 6, a
3
– a
2
= 17 – 11 = 6, a
4
– a
3
= 23 – 17 = 6
As (a
k + 1
– a
k
) is the same for k = 1, 2, 3, etc., the given list of numbers is an AP.
Now, for this AP we have a = 5 and d = 6.
We choose to begin with the assumption that 301 is a term, say, the nth term of the this AP.
We will see if an ‘n’ exists for which a
n
= 301.
We know
a
n
= a + (n – 1) d
So, for 301 to be a term we must have
301 = 5 + (n – 1) × 6
or 301 = 6n – 1
So, n =
302 151
6 3

But n should be a positive integer (Why?).
So, 301 is not a term of the given list of numbers.
Example-7. How many two-digit numbers are divisible by 3?
Solution : The list of two-digit numbers divisible by 3 is :
12, 15, 18, . . . , 99
Is this an AP? Yes it is. Here, a = 12, d = 3, a
n
= 99.

141. Mathematic1 3 8
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As a
n
= a + (n – 1) d,
we have99 = 12 + (n – 1) × 3
i.e., 87 = (n – 1) × 3
i.e., n – 1 =
87
29
3

i.e., n = 29 + 1 = 30
So, there are 30 two-digit numbers divisible by 3.
Example-8. Find the 11
th
term from the last of the the AP series given below :
AP : 10, 7, 4, . . ., – 62.
Solution : Here, a = 10, d = 7 – 10 = – 3, l = – 62,
where l = a + (n – 1) d
To find the 11th term from the last term, we will find the total number of terms in the AP.
So, – 62 = 10 + (n – 1)(–3)
i.e., – 72 = (n – 1)(–3)
i.e., n – 1 = 24
or n = 25
So, there are 25 terms in the given AP.
The 11
th
term from the last will be the 15
th
term of the series. (Note that it will not be the
14
th
term. Why?)
So,a
15
= 10 + (15 – 1)(–3) = 10 – 42 = – 32
i.e., the 11
th
term from the end is – 32.
Example-9. A sum of Rs 1000 is invested at 8% simple interest per year. Calculate the interest
at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30
years making use of this fact.
Solution : We know that the formula to calculate simple interest is given by
Simple Interest =
P R T
100
 
So, the interest at the end of the 1
st
year = Rs.
1000 8 1
100
 
= Rs 80
The interest at the end of the 2
nd
year = Rs
1000 8 2
100
 
= Rs 160

142. Progressions1 3 9
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The interest at the end of the 3
rd
year =
1000 8 3
100
 
= Rs 240
Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on. So, the
interest (in Rs) at the end of the 1
st
, 2
nd
, 3
rd
, . . . years, respectively are
80, 160, 240, . . .
It is an AP as the difference between the consecutive terms in the list is 80,
i.e., d = 80. Also, a = 80.
So, to find the interest at the end of 30 years, we shall find a
30
.
Now, a
30
= a + (30 – 1) d = 80 + 29 × 80 = 2400
So, the interest at the end of 30 years will be Rs 2400.
Example-10. In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in
the third, and so on. There are 5 rose plants in the last row. How many rows are there in the
flower bed?
Solution : The number of rose plants in the 1
st
, 2
nd
, 3
rd
, . . ., rows are :
23, 21, 19, . . ., 5
It forms an AP (Why?).
Let the number of rows in the flower bed be n.
Thena = 23, d = 21 – 23 = – 2, a
n
= 5
As, a
n
= a + (n – 1) d
We have,5 = 23 + (n – 1)(– 2)
i.e., – 18 = (n – 1)(– 2)
i.e., n = 10
So, there are 10 rows in the flower bed.
EXERCISE - 6.2
1. Fill in the blanks in the following table, given that a is the first term, d the common
difference and a
n
the nth term of the AP:
S. No.a d n a
n
(i) 7 3 8 . . .
(ii) – 18 . . . 10 0

143. Mathematic1 4 0
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(iii). . . – 3 18 – 5
(iv) – 18.9 2.5 . . . 3.6
(v) 3.5 0 105 . . .
2.Find the
(i)30
th
term of the A.P. 10, 7, 4 ......
(ii)11
th
term of the A.P. :
1
3, , 2,.....
2


3.Find the respective terms for the following APs.
(i)a
1
= 2; a
3
= 26 find a
2
(ii)a
2
= 13; a
4
= 3 find a
1
, a
3
(iii)a
1
= 5; a
4
= 9
1
2
find a
2
, a
3
(iv)a
1
= 4; a
6
= 6 find a
2
, a
3
, a
4
, a
5
(v)a
2
= 38; a
6
= 22 find a
1
, a
3
, a
4
, a
5
4.Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
5. Find the number of terms in each of the following APs :
(i)7, 13, 19, . . . , 205(ii)18, 15
1
2

6. Check whether, –150 is a term of the AP : 11, 8, 5, 2 . . .
7. Find the 31
st
term of an AP whose 11
th
term is 38 and the 16
th
term is 73.
8. If the 3
rd
and the 9
th
terms of an AP are 4 and – 8 respectively, which term of this AP is
zero?
9. The 17
th
term of an AP exceeds its 10
th
term by 7. Find the common difference.
10.Two APs have the same common difference. The difference between their 100th terms
is 100, what is the difference between their 1000th terms?
11.How many three-digit numbers are divisible by 7?
12.How many multiples of 4 lie between 10 and 250?
13.For what value of n, are the nth terms of two APs: 63, 65, 67, . .. and 3, 10, 17, . . .
equal?
14.Determine the AP whose third term is 16 and the 7
th
term exceeds the 5
th
term by 12.
15.Find the 20
th
term from the end of the AP : 3, 8, 13, . . ., 253.

144. Progressions1 4 1
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16.The sum of the 4
th
and 8
th
terms of an AP is 24 and the sum of the 6
th
and 10
th
terms is
44. Find the first three terms of the AP.
17.Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment
of Rs 200 each year. In which year did his income reach Rs 7000?
18.Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by
Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.
6.4 SUM OF FIRST n TERMS OF AN ARITHMETIC PROGRESSION
Let us consider the situation again given in Section 6.1 in
which Shakila put Rs 1000 money box when her daughter was
one year old, Rs 1500 on her second birthday, Rs 2000 on her
third birthday and will continue in the same way. How much money
will be collected in the money box by the time her daughter is 21
years old?
Here, the amount of money (in Rs) put in the money box on
her first, second, third, fourth . . . birthday were respectively 1000,
1500, 2000, 2500, . . . till her 21st birthday. To find the total
amount in the money box on her 21st birthday, we will have to
write each of the 21 numbers in the list above and then add them up. Don’t you think it would be
a tedious and time consuming process? Can we make the process shorter?
This would be possible if we can find a method for getting this sum. Let us see.
6.4.1 WHAT GAU SS DID
We consider the problem given to Gauss (about whom you read in Chapter 1), to solve
when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to
100. He immediately replied that the sum is 5050. Can you guess how did he do? He wrote :
S = 1 + 2 + 3 + . . . + 99 + 100
And then, reversed the numbers to write
S = 100 + 99 + . . . + 3 + 2 + 1
When he added these two he got 25 as both the sums have to be equal. So he work,
2S = (100 + 1) + (99 + 2) + . . . + (3 + 98) + (2 + 99) + (1 + 100)
= 101 + 101 + . . . + 101 + 101 (100 times) (check this out and discuss)
So,S =
100 101
2

= 5050, i.e., the sum = 5050.

145. Mathematic1 4 2
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6.4.2 SUM OF n TERM S OF AN AP.
We will now use the same technique to find the sum of the first n terms of an AP :
a, a + d, a + 2d, . . .
The nth term of this AP is a + (n – 1) d.
Let S denote the sum of the first n terms of the A.P. Whose n
th
term is a
n
.
The term before the nth term is (a
n
 d) and the one before
terms
n
S ( ) ( 2 ) ... ( 2 ) ( )
n
n n n
a a d a d a d a d a          

is (a
n
 2d)
Rewriting the terms in reverse order
n
S ( ) ( 2 ) ... ( 2 ) ( )
n n n n
a a d a d a d a d a          
Adding the two like Gauss did we get.
2S
n
= a + a
n
+ (a + a
n
) + (a + a
n
) + ... + (a + a
n
) + (a + a
n
) + (a + a
n
)
2S
n
= n(a + a
n
)
S
n
[2 ( 1) ]
2
n
a n d   [ a
n
= a + (n1)d]
This form of the result is useful when the first and the last terms of an AP are given and the
common difference is not given.
Money for Shakila’s daughter
Now we return to the question that was posed to us in the beginning. The amount of
money (in Rs) in the money box of Shakila’s daughter on 1
st
, 2
nd
, 3
rd
, 4th birthday, . . ., were
1000, 1500, 2000, 2500, . . ., respectively.
This is an AP. We have to find the total money collected on her 21st birthday, i.e., the sum
of the first 21 terms of this AP.
Here, a = 1000, d = 500 and n = 21. Using the formula :
S
n
[2 ( 1) ],
2
n
a n d  
we have S =
21
[2 1000 (21 1) 500]
2
   
=
21
[2000 10000]
2


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Free Distribution by A.P. Government
=
21
[12000] 126000
2

So, the amount of money collected on her 21
st
birthday is Rs 12600.
We use S
n
in place of S to denote the sum of first n terms of the AP so that we know how
many terms we have added. We write S
20
to denote the sum of the first 20 terms of an AP. The
formula for the sum of the first n terms involves four quantities S
n
, a, d and n. If we know any
three of them, we can find the fourth.
Remark : The n
th
term of an AP is the difference of the sum to first n terms and the sum to first
(n – 1) terms of it, i.e., a
n
= S
n
– S
n – 1
.
DO THIS
Find the sum of indicated number of terms in each of the following A.P.s
(i)16, 11, 6 .....; 23 terms(ii)0.5, 1.0, 1.5, .....; 10 terms
(iii)
1 3
1, , .....,
4 2
 10 terms
Let us consider some examples.
Example-11. If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the
20th terms.
Solution : Here, S
n
= 1050; n = 14, a = 10
S
n
= [2 ( 1) ]
2
n
a n d 
1050 =
14
[2 13 ] 140 91
2
a d d  
Example-12. How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?
Solution : Here, a = 24, d = 21 – 24 = –3, Sn = 78. We need to find n.
We know that S
n
=[2 ( 1) ]
2
n
a n d 
So, 78 [48 ( 1( 3)] [51 3 ]
2 2
n n
n n     
or 3n
2
– 51n + 156 = 0
or n
2
– 17n + 52 = 0

147. Mathematic1 4 4
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or (n – 4)(n – 13) = 0
or n = 4 or 13
Both values of n are admissible. So, the number of terms is either 4 or 13.
Remarks :
1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78.
2. Two answers are possible because the sum of the terms from 5
th
to 13
th
will be zero.
This is because a is positive and d is negative, so that some terms are positive and some
are negative, and will cancel out each other.
Example-13. Find the sum of :
(i)the first 1000 positive integers (ii) the first n positive integers
Solution :
(i)Let S = 1 + 2 + 3 + . . . + 1000
Using the formula S
n
= ( )
2
n
a l for the sum of the first n terms of an AP, we have
S
1000
=
1000
(1 1000)
2
 = 500 × 1001 = 500500
So, the sum of the first 1000 positive integers is 500500.
(ii)Let S
n
= 1 + 2 + 3 + . . . + n
Here a = 1 and the last term l is n.
Therefore, S
n
=
(1 )
2
n n
(or) S
n
=
( 1)
2
n n
So, the sum of first n positive integers is given by
S
n
=
( 1)
2
n n
Example-14. Find the sum of first 24 terms of the list of numbers whose nth term is given by
a
n
= 3 + 2n
Solution :Asa
n
= 3 + 2n,
so, a
1
= 3 + 2 = 5
a
2
= 3 + 2 × 2 = 7

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Free Distribution by A.P. Government
a
3
= 3 + 2 × 3 = 9

List of numbers becomes 5, 7, 9, 11, . . .
Here, 7 – 5 = 9 – 7 = 11 – 9 = 2 and so on.
So, it forms an AP with common difference d = 2.
To find S
24
, we have n = 24, a = 5, d = 2.
Therefore,S
24
=
24
[2 5 (24 1) 2] 2[10 46] 672
2
      
So, sum of first 24 terms of the list of numbers is 672.
Example-15. A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the
seventh year. Assuming that the production increases uniformly by a fixed number every year,
find :
(i)the production in the 1
st
year (ii)the production in the 10
th
year
(iii)the total production in first 7 years
Solution : (i) Since the production increases uniformly by a fixed number every year, the number
of TV sets manufactured in 1
st
, 2
nd
, 3
rd
, . . ., years will form an AP.
Let us denote the number of TV sets manufactured in the nth year by a
n
.
Then, a
3
= 600 and a
7
= 700
or, a + 2d = 600
and a + 6d = 700
Solving these equations, we get d = 25 and a = 550.
Therefore, production of TV sets in the first year is 550.
(ii)Now a
10
= a + 9d = 550 + 9 × 25 = 775
So, production of TV sets in the 10
th
year is 775.
(iii)Also, S
7
=
7
[2 550 (7 1) 25]
2
   

7
[1100 150] 4375
2
  
Thus, the total production of TV sets in first 7 years is 4375.

149. Mathematic1 4 6
Free Distribution by A.P. Government
EXERCISE - 6.3
1. Find the sum of the following APs:
(i)2, 7, 12, . . ., to 10 terms.(ii)–37, –33, –29, . . ., to 12 terms.
(iii)0.6, 1.7, 2.8, . . ., to 100 terms.(iv)
1 1 1
, , , .....,to 11terms.
15 12 10
2. Find the sums given below :
(i)
1
7 10 14 .... 84
2
    (ii)34 + 32 + 30 + . . . + 10
(iii)–5 + (–8) + (–11) + . . . + (–230)
3. In an AP:
(i)given a = 5, d = 3, a
n
= 50, find n and S
n
.
(ii)given a = 7, a
13
= 35, find d and S
13
.
(iii)given a
12
= 37, d = 3, find a and S
12
.
(iv)given a
3
= 15, S
10
= 125, find d and a
10
.
(v)given a = 2, d = 8, S
n
= 90, find n and a
n
.
(vi)given a
n
= 4, d = 2, S
n
= –14, find n and a.
(vii)given l = 28, S = 144, and there are total 9 terms. Find a.
4. The first and the last terms of an AP are 17 and 350 respectively. If the common difference
is 9, how many terms are there and what is their sum?
5. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18
respectively.
6. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first
n terms.
7. Show that a
1
, a
2
, . . ., a
n
, . . . form an AP where a
n
is defined as below :
(i)a
n
= 3 + 4n (ii)a
n
= 9 – 5n
Also find the sum of the first 15 terms in each case.
8. If the sum of the first n terms of an AP is 4n – n
2
, what is the first term (remember the first
term is S
1
)? What is the sum of first two terms? What is the second term? Similarly, find
the 3rd, the 10th and the nth terms.
9. Find the sum of the first 40 positive integers divisible by 6.
10.A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their
overall academic performance. If each prize is Rs 20 less than its preceding prize, find
the value of each of the prizes.

150. Progressions1 4 7
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11.In a school, students thought of planting trees in and around the school to reduce air
pollution. It was decided that the number of trees, that each section of each class will
plant, will be the same as the class, in which they are studying, e.g., a section of Class I
will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are
three sections of each class. How many trees will be planted by the students?
12.A spiral is made up of successive semicircles, with
centres alternately at A and B, starting with centre at
A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as
shown in Fig. 5.4. What is the total length of such a
spiral made up of thirteen consecutive semicircles?
(Take
22
7
 )
[Hint : Length of successive semicircles is l
1
, l
2
, l
3
, l
4
, . . . with centres at A, B, A, B, . . .,
respectively.]
16.200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next
row, 18 in the row next to it and so on. In how may rows are the 200 logs placed and
how many logs are in the top row?
17.In a bucket and ball race, a bucket is placed at the starting point, which is 5 m from the
first ball, and the other balls are placed 3 m apart in a straight line. There are ten balls in
the line.
A competitor starts from the bucket, picks up the nearest ball, runs back with it, drops it
in the bucket, runs back to pick up the next ball, runs to the bucket to drop it in, and she
continues in the same way until all the balls are in the bucket. What is the total distance
the competitor has to run?
[Hint : To pick up the first ball and the second ball, the total distance (in metres) run by
a competitor is 2 × 5 + 2 × (5 + 3)]

1

3

2

4
AB
5 m3 m3 m

151. Mathematic1 4 8
Free Distribution by A.P. Government
6.5 GEO METRIC PROG RESSIONS
Consider the lists
(i)30, 90, 270, 810 ..... (ii)
1 1 1 1
, , , .....
4 16 64 256
(iii)30, 24, 19.2, 15.36, 12.288
Given a term, can we write the next term in each of the lists above ?
in (i) each term is obtained by multiplying the preceeding term by 3.
in (ii) each term is obtained by multiplying the preceeding term by
1
4
.
in (iii) each term is obtained by multiplying the preceeding term by 0.8.
In all the lists above, we see that successive terms are obtained by multiplying the preceeding
term by a fixed number. Such a list of numbers is said to form Geometric Progression (GP).
This fixed number is called the common ration ‘r’ of G.P. So in the above example (i), (ii), (iii) the
common ratios are 3,
1
4
, 0.8 respectively..
Let us denote the first term of a GP by a and common ratio r. To get the second term
according to the rule of Geometric Progression, we have to multiply the first term by the common
ratio r.
The second term = ar
a, ar, ar
2
.....
This is called the general form of a GP.
in the above GP the ratio between any term (except first term) and its preceding term is ‘r’
i.e.,
2
..........
ar ar
r
a ar
  
If we denote the first term of GP by a
1
, second term by a
2
..... nth term by an
then
32
1 2 1
...... 2
n
n
a aa
a a a

   
A list of numbers a
1
, a
2
, a
3
.... a
n
... is called a geometric progression (GP), if each
term is non zero and
1
n
n
a
r
a



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Wheren natural number.
r real number 
DO THIS
Find which of the following are not G.P.
1.6, 12, 24, 48, ..... 2.1, 4, 9, 16, ......
3.1, 1, 1, 1, ..... 4.4, 20, 100, 500, .....
Some more example of GP are :
(i)A person writes a letter to four of his friends. He asks each one of them to copy the letter
and give it to four different persons with same instructions so that they can move the
chain ahead similarly. Assuming that the chain is not broken the number of letters at first,
second, third ... stages are
1, 4, 16, 256 .............. respectively.
(ii)The total amount at the end of first , second, third .... year if ` 500/- is deposited in the
bank with annual rate 10% interest compounded annually is
550, 605, 665.5 ......
(iii)A square is drawn by joining the mid points of the sides
of a given square. A third square is drawn inside the
second square in the same way and this process
continues indefinitely. If a side of the first square is
16cm then the area of first, second, third ..... square
will be respectively.
256, 128, 64, 32, .....
(iv)Initially a pendutum swings through an arc of 18 inches.
On each sucessive swing the length of the arc is 0.9 of the
previous length. So the length of the arc at first, second, third.......
swing will be resepectively.
18, 16.2, 14.58, 13.122......
THINK AND DISCUSS
1. Explain why each of the lists above is a G.P.
2. To know about a G.P. what is minimum information that you need ?

153. Mathematic1 5 0
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Now let us term how to construct a G.P. when the first term ‘a’ and common ratio ‘r’ are
given. And also learn how to check whether the given list of numbers is a G.P.
Example-16. Write the G.P. if the first term a = 3, and the common ratio r = 2.
Solution : Since ‘a’ is the first term it can easily be written
We know that in G.P. every succeeding term is obtained by multiplying the preceding
term with common ratio ‘r’. So to get the second term we have to multiply the first term a = 3
by the common ratio r = 2.
 Second term = ar = 3 ×2 = 6
Similarly the third term = second term × common ratio
= 6 × 2 = 12
If we proceed in this way we get the following G.P.
3, 6, 12, 24,..... .
Example-17. Write G.P. if a = 256, r =
1
2

Solution : General form of GP = a, ar, ar
2
, ar
3
, .....
= 256, 256
1
2
 
 
 
, 256
2
1
2
 
 
 
, 256
3
1
2
 
 
 
= 256, 128, 64, 32 ......
Example-18. Find the common ratio of the GP 25, 5, 1,
1
5

.
Solution : We know that if the first, second, third .... terms of a GP are a
1
, a
2
, a
3
.... respectively
the common ratio
32
1 2
.....
aa
r
a a
  
Here a
1
= 25, a
2
=5, a
3
= 1.
So common ratio
5 1 1
25 5 5
r
 
  

.
Example-19. Which of the following list of numbers form G.P. ?
(i)3, 6, 12, ..... (ii)64, 32, 16,
(iii)
1 1 1
, , ,......
64 32 8

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Solution : (i)We know that a list of numbers a
1
, a
2
, a
3
, .....a
n
..... is called a GP if each term is
non zero and
32
1 2 1
.....
n
n
a aa
r
a a a

  
Here all the terms are non zero. Further
2
1
6
2
3
a
a
  and
3
2
12
2
6
a
a
 
i.e.,
32
1 2
2
aa
a a
 
So, the given list of number form a G.P. which contain ratio 2.
(ii)All the terms are non zero.
2
1
32 1
64 2
a
a
 
 
and
3
1
16 1
32 2
a
a

 

32
1 2
1
2
aa
a a

  
So, the given list of numbers form a GP with common ratio
1
2

.
(iii)All the terms are non zero.
2
1
1
32
2
1
64
a
a
 
3
2
1
8
4
1
32
a
a
 
Here
32
1 2
aa
a a

So, the given list of numbers does not form G.P.

155. Mathematic1 5 2
Free Distribution by A.P. Government
EXERCISE - 6.4
1. In which of the following situations, does the list of numbers involved make a G.P.?
(i)Salary of Hema, when her salary is 5,00,000/- for the first year and expected to
receive yearly increase of 10% .
(ii)Number of bricks needed to make each step, if the stair case has total 30 steps.
Bottom step needs 100 bricks and each successive step needs 2 brick less than
the previous step.
(iii)Perimeter of the each triangle,
when the mid points of sides of an
equilateral triangle whose side is
24 cm are joined to form another
triangle, whose mid points in turn
are joined to form still another
triangle and the process continues
indefinitely.
2. Write three terms of the G.P. when the first term ‘a’ and the common ratio ‘r’ are given?
(i)a = 4 r = 3 (ii) 5a
1
5
r
(iii)a = 81r =
1
3

(iv)
1
64
a r = 2
3. Which of the following are G.P. ? If they are G.P. Write three more terms?
(i)4, 8, 16 ..... (ii)
1 1 1
, , .....
3 6 12

(iii)5, 55, 555, .... (iv)2, 6, 18 .....
(v)
1 1 1
, , .....
2 4 6
(vi)3, 3
2
, 3
3
, .....
(vii)
1
, 1, ,.....x
x
(viii)
1 8
, 2, .....
2 2

(ix)0.4, 0.04, 0.004, .....
4. Find x so that x, x + 2, x + b are conseecitive terms of a geometric progression.
24 24
24

156. Progressions1 5 3
Free Distribution by A.P. Government
6.6 n
th
TERM OF A G.P.
Let us consider the situation (iii) again given in the introduction in which the number of
bacteria in a certain culture triples every hour. If there were 30 bacteria present in the culture
originally. Then, what would be number of bacteria in fourth hour?
To answer this let us first see what the number of bacteria in second hour would be.
Since for every hour it triples
No.of bacteria in Second hour= 3 × no.of bacteria in first hour
= 3 × 30 = 30 × 3
1
= 30 × 3
(2-1)
= 90
No.of bacteria in third hour= 3 × no.of bacteria in second hour
= 3 × 90= 30 × (3×3)
= 30 × 3
2
= 30 × 3
(3-1)
= 270
No.of bacteria in fourth hour = 3 × no.of bacteria in third hour
= 3 × 270= 30 × (3×3×3_
= 30 × 3
3
= 30 × 3
(4-1)
= 810
Observe that we are getting a list of numbers
30, 90, 270, 810, ....
These numbers are in GP (why ?)
Now looking at the pattern formed above, can you find number of bacteria in 20th hour?
You may have already got some idea from the way we have obtained the number of
bacteria as above. By using the same pattern, we can compute that Number of bacteria in 20
th
hour.
19 terms
30 (3 3 ... 3)    

= 30 × 3
19
= 30 × 3
(20-1)
This example would have given you some idea about how to write the 25th term. 35
th
term and more generally the nth term of the G.P.
Let a
1
, a
2
, a
3
..... be in G.P. whose ‘first term’ a
1
is a and the common ratio is r

157. Mathematic1 5 4
Free Distribution by A.P. Government
then the second term a
2
= ar = ar
(2-1)
the third terma
3
= a
2
×r = (ar) × r = ar
2
= ar
(3-1)
the fourth terma
4
= a
3
×r = ar
2
× r = ar
3
= ar
(4-1)
................................................
................................................
Looking at the pattern we can say that n
th
term a
n
= ar
n-1
So n
th
term an of a GP with first term ‘a’ and common ratio ‘r’ is given by a
n
= ar
n-1
.
Let us consider some examples
Example-20. Find the 20
th
and n
th
term of the G.P.
5 5 5
, , .......
2 4 8
Solution : Here a =
5
2
and
5
1
4
52
2
r 
Then
19
20 1
20 10
5 1 5
2 2 2
a ar
  
  
 
 
and
1
15 1 5
2 2 2
n
n
n n
a ar

  
  
 
 
Example-21. Which term of the G.P. : 2, 2 2, 4 ..... is 128 ?
Solution : Here a = 2 r =
2 2
2
2

Let 128 be the n
th
term of the G.P.
Thena
n
= ar
n1
= 128
1
2.( 2) 128
n

1
( 2) 64
n

1
62
(2) 2
n


158. Progressions1 5 5
Free Distribution by A.P. Government

1
6
2
n

 n = 13.
Hence 128 is the 13
th
term of the G.P.
Example-22. In a GP the 3
rd
term is 24 and 65
th
term is 192. Find the 10
th
term.
Solution : Here a
3
= ar
2
= 24 ...(1)
a
6
= ar
5
= 195 ...(2)
Dividing (2) by (1) we get
5
2
195
24
ar
ar

 r
3
= 8 = 2
3
 r = 2
Substituting r = 2 in (1) we get a = 6.
 a
10
= ar
9
= 6(2)
9
= 3072.
EXERCISE-6.5
1. For each geometric progression find the common ratio ‘r’, and then find a
n
(i)
3 3 3
3, , , .........
2 4 8
(ii)2, 6, 18, 54
(iii)1, 3, 9, 18 .... (iv)
4 8
5, 2, , .........
5 25
2. Find the 10
th
and n
th
term of G.P. : 5, 25, 125, .....
3. Find the indicated term of each geometric Progression
(i)a
1
= 9;
1
;
3
r find a
7
(ii)a
1
= 12;
1
;
3
r find a
6
4. Which term of the G.P.
(i)2, 8, 32, ..... is 512 ?(ii)3, 3, 3 3...... is 729 ?
(iii)
1 1 1
, , .....
3 9 27
is
1
2187
?

159. Mathematic1 5 6
Free Distribution by A.P. Government
5. Find the 12
th
term of a G.P. whose 8
th
term is 192 and the common ratio is 2.
6. The 4th term of a geometric progression is
2
3
and the seventh term is
16
81
. Find the
geometric series.
7. If the geometric progressions 162, 54, 18 ..... and
2 2 2
, , ....
81 27 9
have their n
th
term
equal. Find the value of n.
OPTIONAL EXERCISE
[This exercise is not meant for examination]
1. Which term of the AP : 121, 117, 113, . . ., is its
first negative term?
[Hint : Find n for a
n
< 0]
2. The sum of the third and the seventh terms of an
AP is 6 and their product is 8. Find the sum of
first sixteen terms of the AP.
3. A ladder has rungs 25 cm apart. The rungs
decrease uniformly in length from 45 cm at the
bottom to 25 cm at the top. If the top and the
bottom rungs are
1
2
2
m apart, what is the length
of the wood required for the rungs?
[Hint : Number of rungs =
250
1
25
]
4.The houses of a row are numbered consecutively
from 1 to 49. Show that there is a value of x such
that the sum of the numbers of the houses preceding
the house numbered x is equal to the sum of the numbers of the houses following it. Find
this value of x.
[Hint : S
x – 1
= S
49
– S
x
]
5.A small terrace at a football ground comprises of 15 steps each of which is 50 m long and
built of solid concrete.
25 cm
45 cm
1
2
2
m
25 cm

160. Progressions1 5 7
Free Distribution by A.P. Government
Each step has a rise of
1
4
m and a tread of
1
2
m. (see Fig. 5.8). Calculate the total volume
of concrete required to build the terrace.
[Hint : Volume of concrete required to build the first step =
1 1
50
4 2
 m
3
]
* These exercises are not from the examination point of view.
6. 150 workers ware engaged to finish a piece of work in a certain number of days. Four
workers dropped from the work in the second day. Four workers dropped in third day
and so on. It took 8 more days to finish the work. Find the number of days in which the
was and completed.
[let the no.of days to finish the work is ‘x’ then
8
150 [2 150 ( 8 1)( 4)]
2
x
x x

     
[Ans. x = 17  x + 8 = 17 + 8 = 25]
7. A machine costs Rs. 5,00,000/-. If the value depreciates 15% in the first year, 13
1
2
%
in the second year, 12% in the third year and so on. What will be its value at the end of
10 years, when all the percentages will be applied to the original cost?
[Total depreciation = 15 + 13
1
2
+12+....10 terms.
10
S [30 13.5] 82.5%
2
n
  
 after 10 year original cost =100  82.5 = 17.5 i.e., 17.5% of 5,00,000
1
m
4
1
m
2
5 0 m

161. Mathematic1 5 8
Free Distribution by A.P. Government
WHAT WE HAVE DISCUSSED
In this chapter, you have studied the following points :
1. An arithmetic progression (AP) is a list of numbers in which each term is obtained by
adding a fixed number d to the preceding term, except the first term. The fixed number d
is called the common difference.
The general form of an AP is a, a + d, a + 2d, a + 3d, . . .
2. A given list of numbers a
1
, a
2
, a
3
, . . . is an AP, if the differences a
2
– a
1
, a
3
– a
2
, a
4
– a
3
,
. . ., give the same value, i.e., if a
k + 1
– a
k
is the same for different values of k.
3. In an AP with first term a and common difference d, the nth term (or the general term) is
given by a
n
= a + (n – 1) d.
4. The sum of the first n terms of an AP is given by :
S [2 ( 1) ]
2
n
a n d  
5. If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is
given by :
S ( )
2
n
a l  .
6. If a, b, c are in A.P. then
2
a c
b

 and b is called the arithmetic mean of a and c.
7. A Geometric Progression (GP) is a list of numbers in which each term is obtained by
multiplying preceeding term with a fixed number ‘r’ except are first term. This fixed
number is called common ratio ‘r’.
The general form of GP is a, ar, ar
2
, ar
3
....
8. If the first term and common rate of a G.P. are a, r respectively then nth term a
n
= ar
n-1
.

163. 7.1 INTRO DUCTIO N
You know that in chess, the Knight moves in ‘L’ shape or two and a half steps (see figure).
It can jump over other pieces too. A Bishop moves diagonally, as many steps as are free in front
of it.
Find out how other pieces move. Also locate Knight,
Bishop and other pieces on the board and see how they
move.
i.Consider that the Knight is at the origin (0, 0). It can
move in 4 directions as shown by dotted lines in the
figure. Find the coordinates of its position after the
various moves shown in the figure.
ii.Find the distance covered by the Knight in each of
its 8 moves i.e. find the distance of A, B, C, D, E, F,
G and H from the origin.
7.2 DISTANCE BETWEEN TWO POINTS
If two points lie on the X-axis (2,
0) and (6, 0) as shown in figure.
It is easy to find the distance
between two points A and B as 4 units.
We can say the distance between
points lying on X-aixs to the difference
between the x-coordinates.
What is the distance between
(2, 0) and (6, 0)?
The difference in the value of
x-coordinates is
6)  (2) = 4 (Negative)
A B
C
D
EF
G
H
-9
Y
O
X
1
Y
1
X
-8-7-6-5-4-3-2-1 123456789
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
A(2, 0)B(6, 0)
Coordinate Geometry7

164. Mathematic1 6 0
Free Distribution by A.P. Government
We never say the distance in negative values.
So, we calculate that absolute value of the distance.
Therefore, the distance
= |  6)  (2)| = |4| = 4
So, in general for the points
A(x
1
, 0), B(x
2
, 0) on the X-axis.
The distance betweem A and B |x
2
 x
1
|
Similarly, if two points lie on
Y-axis, then the distance between the
points A and B would be the difference
between the y coordinates of the points.
The distance between two points
(0, y
1
) (0, y
2
) would be |y
2
 y
1
|.
For example. Let the points
A(0, 2) and B(0, 7)
The distance between A and B is
|7  2| = 5.
TRY THESE
1.Where do these following points lie (0, 3), (0, 8), (0, 6), (0, 4)
2.What is the distance between (0, 3), (0, 8) and justify that the distance between two
points on Y-axis is |y
2
 y
1
|.
THINK AND DISCUSS
How will you find the distance between two points in two different quadrants?
7.3 DISTANCE BETWEEN TWO POINTS ON A LINE PARALLEL TO COORDINATE
AXIS.
Consider the points A(x
1
, y
1
) and B(x
2
, y
2
). Since the y-coordinates are equal. So points
lie on a line, parallel to X-axis.
Draw AP and BQ perpendicular to X-axis.
Observe the figure. The distance between two points A and B is equal to the distance
between P and Q.
-9
Y
O
X
1
Y
1
X
-8-7-6-5-4-3-2-1 123456789
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
A (0, 2)
B (0, 7)

165. Coordinate Geometry1 6 1
Free Distribution by A.P. Government
Therefore,
Distance AB = Distance
PQ = |x
2
 x
1
| (i.e., The difference
between x coordinates)
Example-1. What is the distance between A (4,0) and B (8, 0).
Solution :The difference in the x coordinates is |x
2
- x
1
| = |8  4| = 4 units.
Example-2. A and B are two points described in (8, 3), (4, 3). Find the distance between A
and B.
Solution : Here x
1
and x
2
are lying in two different quadrant and y-coordinate are equal.
Distance AB = |x
2
 x
1
| = |4  8| = |12| = 12 units
DO THIS
Find the distance between the
following points which are lying on the
lines parallel to the X-axis.
i. (3, 8), (6, 8) ii. (4,3), (8,3)
iii. (3, 4), (6, 4)
Let A and B denote the points
(4, 0) and (0, 3) and ‘O’ be the
origin.
The AOB is a right angle triangle.
From the figure
-9
Y
O
X
1
Y
1
X
-8-7-6-5-4-3-2-1 12P456Q89
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
A ( , )x y
1 1 B ( , )x y
2 2
( - x x
2 1)
-9
Y
O
X
1
Y
1
X
-8-7-6-5-4-3-2-1 123456789
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
A (4, 0)
B (0, 3)

166. Mathematic1 6 2
Free Distribution by A.P. Government
OA = 4 units (x-coordinate)
OB = 3 units (y-coordinate)
Then distance AB = ?
Hence, by using pythagoran theorem
AB
2
= AO
2
+ OB
2
AB
2
= 4
2
+ 3
2
AB = 16 9 25 5    is the distance between A and B.
TRY THIS
Find the distance between points ‘O’ (origin) and ‘A’ (7, 4).
7.4 DISTANCE BETWEEN ANY TWO POINTS (ON A LINE) IN A PLANE
Let A(x
1
, y
1
) and B(x
2
, y
2
) be
any two points (on a line) in a plane as
shown in figure.
Draw AP and BQ perpendiculars
to X-axis
Draw a perpendicular from the
point A on BQ to meet at the point R is
AR.
Then OP = x
1
, OQ = x
2
So PQ = OQ  OP = x
2
 x
1
Observe the shape of APQR. It is
a rectangle.
So PQ = AR = x
2
 x
1
.
Also QB = y
2
, QR = y
1
,
So BR = QB  QR = y
2
 y
1
from ARB (right angle triangle)
AB
2
= AR
2
+ RB
2
(By Pythagoras theorem)
AB
2
= (x
2
 x
1
)
2
+ (y
2
 y
1
)
2
i.e., AB =
2 2
2 1 2 1
( ) ( )x x y y  
Hence, the distance between the points A and B is
-9
Y
O
X
1
Y
1
X
-8-7-6-5-4-3-2-1 12P456Q89
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
A ( , )x y
1 1
B ( , )x y
2 2
( - x x
2 1)
R
( - x x
2 1)
(y -
2 1y)

167. Coordinate Geometry1 6 3
Free Distribution by A.P. Government
d =
2 2
2 1 2 1
( ) ( )x x y y  
Which is called the distance formulas.
THINK AND DISCUSS
1.Ramu says the distance of a point P(x
1
, y
1
) from the origin O(0, 0) is
2 2
x y. Are you
agree with Ramu or not? Why?
2.Ramu also writes the distance formulas as
AB =
2 2
2 1 2 1
( ) ( )x x y y   (why?)
Example-3.Let’s find the distance between two point A(4, 2) and B(8, 6)
Solution : Compare these points with (x
1
, y
1
), (x
2
, y
2
)
x
1
= 4, x
2
= 8, y
1
= 3, y
2
= 6
Using distance formula
distance AB = d =
2 2
2 1 2 1
( ) ( )x x y y  
=
2 2 2 2
(8 4) (6 3) 4 3    
16 9 25   = 5 units.
DO THIS
Find the distance between the following points
(i)(7, 8) and (2, 3) (ii)(8, 6) and (2, 0)
TRY THIS
Line AB has end points A(1, 3) and B(4, 4), find AB to the nearest tenth.
THINK AND DISCUSS
Sridhar calculated the distance between T(5, 2) and R(4, 1) to the nearest tenth is
9.5 units.
Now you find the distance between P (4, 1) and Q (-5, -2). Do you get the same answer
that sridhar got? Why?

168. Mathematic1 6 4
Free Distribution by A.P. Government
Let us see some examples
Example-4. Show that the points A (4, 2), B (7, 5) and C (9, 7) are three point lie on a same
line.
Solution : Let us assume condition is that AB + BC = AC and lie on same line as give figure.
Now, we find the distance of AB, BC, AC
By using distance formula =
2 2
2 1 2 1
( ) ( )d x x y y   
So, d = AB =
2 2 2 2
(7 4) (5 2) 3 3 9 9 18       
9 2 3 2   units.
BC =
2 2 2 2
(9 7) (7 5) 2 2 4 4 8 2 2         units
AC =
2 2 2 2
(9 4) (7 2) 5 5 25 25 50       
25 2 5 2   units.
Now AB + BC
=3 2 2 2 5 2  = As so the condition is satisfied, that the three points (4, 2), (7, 5)
and (9, 7) are lie on a straight line. (Point they lie on same line called as collinear point).
Example-5. Is the points (3, 2), (2, 3) and (2, 3) form a triangle?
Solution : Let us apply the distance formuls to find the distances PQ, QR and PR, where P(3,
2), Q(2, 3) and R(2, 3) are the given points. We have
PQ =    
2 2 2 2
2 3 3 2 5 5 25 25 50            = 7.07 (approx)
QR =     
2 2 2 2
2 2 3 3 4 6 52 7.21         (approx)
PR =  
2 2 2 2
2 3 3 2 1 1 2 1.41        (approx)
Since the sum of any two of these distances is greater than the third distance, therefore, the
points P, Q and R form a triangle and all the sides of triangle is unequal.
Example-6. Show that the points (1, 7), (4, 2), (1, 1) and (4, 4) are the vertices of a
square.

169. Coordinate Geometry1 6 5
Free Distribution by A.P. Government
Solution : Let A(1, 7), B(4, 2), C(1, 1)and D(4, 4) be the given points.
One way of showing that ABCD is a square is to use the property that all its sides should
be equal. And both its digonals should also be equal. Now
So sides areAB = d =
2 2
(1 4) (7 2) 9 25 34     
BC =
2 2
(4 1) (2 1) 25 9 34     
CD =
2 2
( 1 4) ( 1 4) 9 25 34       
DA =  
2 2
4 1 4 7    = 25 9 = 34
and digonal areAC =
2 2
(1 1) (7 1) 4 64 68     
BD =
2 2
(4 4) (2 4) 64 4 68     
Since AB = BC = CD = DA and AC = BD. So all the four sides of the quadrilateral
ABCD are equal and its diagonals AC and BD are also equal. Therefore, ABCD is square.
Example-7. Figure show the arrangement of desks
in a class room.
Madhuri, meena, pallavi are seated at A(3, 1),
B(6, 4) and C(8, 6) respectively.
Do you think they are seated in a line ?
Give reasons for your answer.
Solution : Using the distance formula, we have
AB=
2 2
6 3 4 1 = 9 9 = 18 = 3 2   
BC= 
2 2
18 6 6 4 = 4 4 = 8 = 2 2   
AC=
2 2
8 3 6 1 = 25 25 = 50 = 5 2   
Since, AB + BC = 3 2 + 2 2 + 5 2 = AC, we can say that the points A, B and C are
collinear. Therefore, they are seated in a line.
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170. Mathematic1 6 6
Free Distribution by A.P. Government
Example-8. Find a relation between x and y such that the point (x , y) is equidistant from the
points (7, 1) and (3, 5).
Solution : Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).
Given that AP = BP. So, AP
2
= BP
2
i.e., (x – 7)
2
+ (y – 1)
2
= (x – 3)
2
+ (y – 5)
2
i.e., (x
2
– 14x + 49) + (y
2
– 2y + 1) = (x
2
– 6x + 9) + (y
2
– 10y + 25)
(x
2
+ y
2
- 14x - 2y + 50) - (x
2
+ y
2
- 6x - 10y + 34) = 0
-8x + 8y = -16
i.e., x – y = 2
Which is the required relation.
Example-9. Find a point on the y-axis which is equidistant from the points A(6, 5) and
B(– 4, 3).
Solution : We know that a point on the Y-axis is of the form (0, y). So, let the point
P(0, y) be equidistant from A and B. Then
PA = 
2 2
6 0 5 y  
PB =  
2 2
4 0 3 4   
PA
2
= PB
2
So,(6 – 0)
2
+ (5 – y)
2
= (– 4 – 0)
2
+ (3 – y)
2
i.e., 36 + 25 + y
2
– 10y = 16 + 9 + y
2
– 6y
i.e., 4y = 36
i.e., y = 9
So, the required point is (0, 9).
Let us check our solution :AP =
2 2
(6 0) (5 9) 36 16 52     
BP =
2 2
( 4 0) (3 9) 16 36 52      
So (0, 9) is equidistant from (6, 5) and (4, 3).

171. Coordinate Geometry1 6 7
Free Distribution by A.P. Government
EXERCISE 7.1
1.Find the distance between the following pairs of points
(i)(2, 3) and (4, 1) (ii)(5, 7) and (1, 3)
(iii)(2, 3) and (3, 2) (iv)(a, b) and (a, b)
2.Find the distance between the points (0, 0) and (36, 15).
3.Show that the points (1, 5), (2, 3) and (2, 1) are collinear.
4.Check whether (5, 2), (6, 4) and (7, 2) are the vertices of an isosceles triangle.
5.In a class room, 4 friends are seated at the points A,
B, C and D as shown in Figure. Jarina and Phani
walk in to the class and after observing for a few
minutes Jarina asks phani “Don’t you think ABCD
is a square?” Phani disagrees, using distance formula,
find which of them correct. Why ?
6.Show that the following points form a equilateral
triangle A(a, 0), B(a, 0), C(0, a3)
7.Prove that the points (7, 3), (5, 10), (15, 8) and (3, 5) taken in order are the corners
of a parallelogram.
8.Show that the points (4, 7), (1, 2), (8, 5) and (5, 4) taken inorder are the vertices of
a rhombus.
9.Name the type of quadrilateral formed, if any, by the following points, and give reasons for
your answer.
(i)(1, 2), (1, 0), (1, 2), (3, 0) (ii)(3, 5), (3, 1), (0, 3), (1, 4)
(iii)(4, 5), (7, 6), (4, 3), (1, 2)
10.Find the point on the x-axis which is equidistant from (2, 5) and (2, 9).
11.If the distance between two points (x, 7) and (1, 15) is 10, find the value of x.
12.Find the values of y for which the distance between the points P(2, 3) and Q(10, y) is 10
units.
13.Find the radius of the circle whose centre is (3, 2) and passes through (5, 6).
14.Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14) ? Give reason.
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172. Mathematic1 6 8
Free Distribution by A.P. Government
7.5 SECTION FORMULA
Suppose a telephone company
wants to position a relay tower at P
between A and B is such a way that the
distance of the tower from B is twice its
distance from A. If P lies on AB, it will
divide AB in the ratio 1 : 2 (See figure). If
we take A as the origin O, and 1 km as
one unit on both the axis, the coordinates
of B will be (36, 15). In order to know the
position of the tower, we must know the coordinates of P. How do we find these coordinates?
Let the coordinates of P be (x, y). Draw perpendiculars from P and B to the x-axis, meeting it in
D and E, respectively. Draw PC perpendicular to BE. Then, by the AA similarity criterion,
studied in Chapter 6, POD and BPC are similar.
Therefore,
OD OP 1
PC PB 2
  and
PD OP 1
BC PB 2
 
So,
1
36 2
x
x


1
15 2
y
y


.
2x = (36  x) 2y = 15  y
3x = 36 3y = 15
x = 12 y = 5
These equations give x = 12
and y = 5.
You can check that P(12, 5)
meets the condition that OP : PB =
1 : 2.
Consider any two points
A(x
1
, y
1
) and B(x
2
, y
2
) and assume
that P (x, y) divides AB internally in
the ratio m
1
: m
2
, i.e.,
1
2
PA
PB
m
m

(See figure).
Draw AR, PS and BT
perpendicular to the x-axis. Draw
Y
36-x
36 km
15 km
B(36, 15)
C
P
y
y
xA
36-x
D
( , )x y
O
X
Y
R
B( )x y
2 2
,
C
P
A
Q
( )x y
1 1
,
O
X
S
m
1
m
2( , )x y

173. Coordinate Geometry1 6 9
Free Distribution by A.P. Government
AQ and PC parallel to the X-axis. Then, by the AA similarity criterion,
PAQ ~ BPC
Therefore,
PA AQ PQ
BP PC BC
  .....(1)
Now,AQ = RS = OS – OR = x – x
1
PC = ST = OT – OS = x
2
– x
PQ = PS – QS = PS – AR = y – y
1
BC = BT– CT = BT – PS = y
2
– y
Substituting these values in (1), we get
1 1 1
2 2 2
m x x y y
m x x y y
 
 
 
Taking
1 1
2 2
m x x
m x x



, we get
1 2 2 1
1 2
m x m x
x
m m



Similarly, taking
1 1
2 2
,
m y y
m y y



we get
1 2 2 1
1 2
m y m y
y
m m



So, the coordinates of the point P(x, y) which divides the line segment joining the points
A(x
1
, y
1
) and B(x
2
, y
2
), internally, in the ratio m
1
: m
2
are
1 2 2 1 1 2 2 1
1 2 1 2
,
m x m x m y m y
m m m m
  
 
 
 
.....(2)
This is known as the section formula.
This can also be derived by drawing perpendiculars from A, P and B on the y-axis and
proceeding as above.
If the ratio in which P divides AB is k : 1, then the coordinates of the point P will be
2 1 2 1
, .
1 1
kx x ky y
k k
  
 
  
Special Case : The mid-point of a line segment divides the line segment in the ratio 1 : 1.
Therefore, the coordinates of the mid-point P of the join of the points A(x
1
, y
1
) and B(x
2
, y
2
) is
1 2 1 2 1 2 1 2
1. 1. 1. 1.
, , .
1 1 1 1 2 2
x x y y x x y y      

   
    

174. Mathematic1 7 0
Free Distribution by A.P. Government
Let us solve few examples based on the section formula.
Example-10. Find the coordinates of the point which divides the line segement joining the
points (4, 3) and (8, 5) in the ratio 3 : 1 internally.
Solution : Let P(x, y) be the required point. Using the section formuls
1 2 2 11 2 2 1
1 2 1 2
P( , ) ,
m y m ym x m x
x y
m m m m
 
 
 
 
, we get
3(8) 1(4) 24 4 28
7,
3 1 4 4
x
 
   

3(5) 1( 3) 15 3 12
3
3 1 4 4
y
  
   

P(x, y) = (7, 3) is the required point.
Example-11. Find the mid point of the line segment joining the points (3, 0) and (-1, 4)
Solution : mid point M(x, y) of the line segment joining the points (3, 0) and (-1, 4).
M(x, y) =
1 2 1 2
,
2 2
x x y y  
 
 
 mid point of the line segment joining the points (3, 0) and (1, 4) is
M(x, y) =
3 ( 1) 0 4 2 4
, , (1,2)
2 2 2 2
     
 
   
   
.
DO THIS
1Find the point which divides the line segment joining the points (3, 5) and (8, 10) internally
in the ratio 2 : 3
2.Find the midpoint of the line segement joining the points (2, 7) and (12, 7).
TRY THIS
Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ABC
1.The median from A meets BC at D. Find the coordinates of the point D.
2.Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
3.Find the coordinates of points Q and R on medians BE and CF.
A
B C
F
DE
P

175. Coordinate Geometry1 7 1
Free Distribution by A.P. Government
4.Find the points which are dividing the line BE in the ratio 2 : 1 and also dividing the line CF
in the ratio 2 : 1.
5.What do you observe ?
Justify the point devides each median in the ration 2 : 1 is the centriod of a triangle.
7.6 CENTROID OF A TRIANGLE
The centroid of a triangle is the point of
intersection of its medians.
Let A(x
1
, y
1
), B(x
2
, y
2
) and C(x
3
, y
3
) be the
vertices of the triangle ABC.
Let AD be the median bisecting its base.
Then,
2 3 2 3
D ,
2 2
x x y y  

 
 
Now the point G on AD which divides it internally in the ratio 2 : 1, is the centroid.
If (x, y) are the coordinates of G, then
G(x, y) =
2 3 2 3
1 1
2 1( ) 2 1( )
2 2
,
2 1 2 1
x x y y
x y
      
 
    
   
 
  
 
 
=
1 2 3 1 2 3
,
3 3
x x x y y y    
 
 
Hence, the coordinates of the centroid are given by
1 2 3 1 2 3
,
3 3
x x x y y y    
 
 
.
Example-12. Find the centroid of the triangle whose vertices are (3, 5), (7, 4), (10, 2)
respectively.
Solution : The coordinates of the centroid are
1 2 3 1 2 3
,
3 3
x x x y y y    

 
 
A
B C
2
D
1
( ,
1
)yx
1
( )x y
2
,
2
( )x y
3
,
3

176. Mathematic1 7 2
Free Distribution by A.P. Government
3 ( 7) 10 ( 5) 4 ( 2)
, (2, 1)
3 3
       
 
 
 
 the centroid is (2, 1).
DO THIS
Find the centroid of the triangle whose vertices are (4, 6), (2, 2) and (2, 5) respectively.
TRY THIS
The points (2, 3), (x, y), (3, 2) are vertices of a triangle if the centroid of this triangle is
G(0, 0), find (x, y).
Example-13. In what ratio does the point (– 4, 6) divide the line segment joining the points
A(– 6, 10) and B(3, – 8)?
Solution : Let (– 4, 6) divide AB internally in the ratio m
1
: m
2
. Using the section formula, we get
 
1 2 1 2
1 2 1 2
3 6 8 10
4, 6 ,
m m m m
m m m m
   
   
 
 
.....(1)
We know that if (x, y) = (a, b) then x = a and y = b.
So,
1 2
1 2
3 6
4
m m
m m

 

and
1 2
1 2
8 10
6
m m
m m
 


Now,
1 2
1 2
3 6
4
m m
m m

 

gives us
– 4m
1
– 4m
2
= 3m
1
– 6m
2
i.e.,7m
1
= 2m
2
1
2
2
=
7
m
m
i.e.,m
1
: m
2
= 2 : 7
We should verify that the ratio satisfies the y-coordinate also.

177. Coordinate Geometry1 7 3
Free Distribution by A.P. Government
Now,
1
1 2 2
11 2
2
8 10
8 10
1
m
m m m
mm m
m
 
 



(Dividing throughout by m
2
)

2
8 10
7
2
1
7
  


=
16
10
7
9
7


=
16 70
9
 
=
54
9
= 6
Therefore, the point (4, 6) divides the line segment joining the points A(6, 10) and
B (3, 8) in the ratio 2 : 7.
7.7 TRISECTIONAL POINTS OF A LINE
Example-14.Find the coordinates of the points of trisection (i.e., ponts dividing in three equal
parts) of the line segment joining the points A(2,2) and B(7, 4).
Solution : Let P and Q be the points of trisection of AB i.e., AP=PQ=QB see figure.
Therefore, P divides AB internally in the ratio 1 : 2.
Therefore, the coordinates of P are (by applying the section formula)
1 2 2 11 2 2 1
1 2 1 2
P( , ) ,
m y m ym x m x
x y
m m m m
 
 
 
 
1( 7) 2(2) 1(4) 2( 2)
,
1 2 1 2
    
 
  
i.e., 
7 4 4 4 3 0
, , 1,0
3 3 3 3
      
  
   
   
Now, Q also divides AB internally in the ratio 2:1.
So,the coordinates of Q are
2( 7) 1(2) 2(4) 1( 2)
,
2 1 2 1
    

 
  
i.e., 
14 2 8 2 12 6
, , 4,2
3 3 3 3
      
  
   
   

178. Mathematic1 7 4
Free Distribution by A.P. Government
Therefore, the coordinates of the points of trisection of the line segment are P(-1, 0) and
Q(-4, 2)
Example-15. Find the ratio in which the y-axis divides the line segment joining the points
(5, 6) and (1, 4). Also find the point of intersection.
Solution : Let the ratio be K : 1. Then by the section formula, the coordinates of the point which
divides AB in the ratio K : 1 are
K( 1) 1(5) K( 4) 1( 6)
,
K 1 K 1
     
 
  
i.e.,
K 5 4K 6
,
K 1 K 1
    
 
  
This point lies on the y-axis, and we know that on the y-axis the abscissa is 0.
Therefore,
K 5
0
K 1
 


-K + 5 = 0  K = 5.
So, the ratio is K : 1  5 : 1
Putting the value of K = 5, we get the point of intersection as

5 5 4(5) 6
,
5 1 5 1
    
 
  

20 6
0,
6
  
 
 

26
0,
6
 
 
 

13
0,
3
 
 
 
Example-16. Show that the points A(7, 3), B(6, 1), C(8, 2) and D(9, 4) taken inorder are
vertices of a parallelogram.
Solution : Let the points A(7, 3), B(6, 1), C(8, 2) and D(9, 4) are vertices of a parallelogram.
We know that the diagonals of a parallelogram bisect each other.
 So the midpoints of the diagonals AC and DB should be equal.
Now, we find the mid points of AC and DB by using
1 2 1 2
,
2 2
x x y y  
 
 
formula.
midpoint of AC =
7 8 3 2 15 5
, ,
2 2 2 2
    

   
   
midpoint of DB =
9 6 4 1 15 5
, ,
2 2 2 2
    

   
   
A(7,3)
C(8, 2)
B(6, 1)
D(9, 4)

179. Coordinate Geometry1 7 5
Free Distribution by A.P. Government
Hence, midpoint of AC = midpoint of DB.
Therefore, the points A, B, C, D are vertices of a parallelogram.
Example-17. If the points A(6, 1), B(8, 2), C(9, 4) and D(P, 3) are the vertices of a parallelogram,
taken inorder, find the value of P.
Solution : We know that diagonals of parallelogram bisect each other.
So, the coordinates of the midpoint of AC = Coordinates of the midpoint of BD.
i.e.,
6 9 1 4 8 P 5
, ,
2 2 2 2
     

   
   
15 5 8 P 5
, ,
2 2 2 2
   

   
   
15 8 P
2 2


15 = 8 + P  P = 7.
EXERCISE - 7.2
1.Find the coordinates of the point which divides the join of (1, 7) and (4, 3) in the ratio
2 : 3.
2.Find the coordinates of the points of trisection of the line segment joining (4, 1) and
(2, 3).
3.Find the ratio in which the line segment joining the points (3, 10) and (6, 8)is divided by
(1, 6).
4.If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find
x and y.
5.Find the coordinates of a point A, where AB is the diameter of a circle whose centre is
(2, 3) and B is (1, 4).
6.If A and B are (2, 2) and (2, 4) respectively. Find the coordinates of P such that
AP =
3
7
AB and P lies on the segment AB.
7.Find the coordinates of points which divide the line segment joining A(4, 0) and B(0, 6)
into four equal parts.

180. Mathematic1 7 6
Free Distribution by A.P. Government
8.Find the coordinates of the points which divide the line segment joining A(2, 2) and
B(2, 8) into four equal parts.
9.Find the area of a rhombous if its vertices are (3, 0), (4, 5), (1, 4) and (2, 1) taken in
order. [Hint : Area of rhombus =
1
2
Product of its diagonals]
10.Find the coordinates of the point which divides the line segment joining the points
(a + b, a  b) and (a  b, a + b) in the ratio 3 : 2 internally.
11.Find the coordinates of centroid of the following:
i.(-1, 3), (6, -3) and (-3, 6)ii.(6, 2), (0, 0) and (4, -7)
iii.(1, -1), (0, 6) and (-3, 0)
7.8 AREA OF THE TRIANGLE
Consider the points A(0, 4) and B(6, 0) form a triangle with
origin O on a plane as shown in figure.
What is the area of the AOB?
AOB is right angle triangle and the base is 6 units (i.e., x
coordinate) and height is 4 units (i.e., y coordinate).
Area of AOB =
1
base height
2
 
=
1
6 4 =12
2
  square units.
TRY THIS
Consider the points of triangle are A(2, 8), B(2, 1) and C(9, 0). What is the area of the
AOB?
THINK AND DISCUSS
Let A(x
1
, y
1
), B(x
2
, y
2
), C(x
3
, y
3
).
Then find the area of the following triangles in a plane.
And discuss with your friends in groups about the area of
those triangle.
B(6, 0)6 units
4
u
n
i
t
s
A(0, 4)
X
A( ,
1
)yx
1
Y
B( )x y
2
,
2
C( )x y
3
,
3
X'
Y'
(i)

181. Coordinate Geometry1 7 7
Free Distribution by A.P. Government
Let ABC be any triangle whose vertices
are A(x
1
, y
1
), B(x
2
, y
2
) and C(x
3
, y
3
) . Draw
AP, BQ and CR perpendiculars from A, B
and C respectively to the x-axis.
Clearly ABQP, APRC and BQRC are
all trapezia as shown in figure.
Now from figure, it is clear that
Area of ABC = area of trapezium
ABQP + area of trapezium APRC  area of
trapezium BQRC
We also know that the
Area of trapezium =
1
2
(sum of parallel sides) (distance between them)
Therefore,
Area of ABC =
1 1 1
(BQ AP)QP (AP CR)PR (BQ CR)QR
2 2 2
    
Here from the figure
BQ = y
2
, AP = y
1
, QP = OP  OQ = x
1
 x
2
CR = y
3
, PR = OR  OP = x
3
 x
1
QR = OR  OQ = x
3
 x
2
Therefore,
2 1 1 2 1 3 3 1 3 3 3 2
1 1 1
( ) ( ) ( ) ( ) ( ) ( )
2 2 2
y y x x y y x x y y x x        
1 2 3 2 3 1 3 1 2
1
[ ( ) ( ) ( )]
2
x y y x y y x y y     
Y
X
B A
C
X'
Y'
(ii)
X
Y
A B
C
Y'
X'
(iii)
X
Y
B
C
A
Y'
X'
(iv)
X X
Y
Y
C( )x y
3 3
,
A( , )y
1
x
1
B( )x y
2 2
,
S
T
U
Q P R

182. Mathematic1 7 8
Free Distribution by A.P. Government
Thus, the area of ABC is the numerical value of the expression
1 2 3 2 3 1 3 1 2
1
[ ( ) ( ) ( )]
2
x y y x y y x y y    
Let us do some examples
Example-18. Find the area of a triangle whose vertices are (1, 1), (4, 6) and (3, 5).
Solution : The area of the triangle formed by the vertices A(1, 1), B(4, 6) and C(3, 5), by
using the formula above
= 1 2 3 2 3 1 3 1 2
1
[ ( ) ( ) ( )]
2
x y y x y y x y y    
is given by
1
[1(6 5) ( 4)( 5 1) ( 3)( 1 6)]
2
         
1
[11 16 21] 24
2
   
So the area of the triangle is 24 square units.
Example-19. Find the area of a triangle formed by the points A(5, 2), B(4, 7) and C(7, 4).
Solution : The area of the triangle formed by the vertices A(5, 2), B(4, 7) and C(7, 4) is given
by
 
1
5(7 4) 4( 4 2) 7(2 7)
2
     
 
1 4
55 24 35 2
2 2

     
Since area is a measure, which cannot be negative, we will take the numberical value of -
2 or absolute value i.e., |2| = 2.
Therefore, the area of the triangle = 2 square units.
DO THIS
1.Find the area of the triangle whose vertices are
2.(5, 2) (3, 5) and (5, 1)
3.(6, 6), (3, 7) and (3, 3)

183. Coordinate Geometry1 7 9
Free Distribution by A.P. Government
TRY THIS
Find the area of the square formed by (0, 1), (2, 1) (0, 3) and (2, 1) taken inorder are
as vertices [Hint: divide the square in to two triangles]
THINK AND DISCUSS
Find the area of the triangle formed by the points P(1.5, 3), Q(6, 2) and R(3, 4)
What do you observe?
Can we have a triangle is 0 square units ?
What does it mean?
7.8.1. CO LLINEA RITY
Suppose the points A(x
1
, y
1
), B(x
2
, y
2
) and C(x
3
, y
3
) are lying on a line. Then, they can
not form a triangle. i.e. area of ABC is zero.
When the area of a triangle is zero then the three points said to be collinear points.
Example-20. The points (3, 2) (2, 8) and (0, 4) are three points in a plane. Show that these
points are collinear.
Solution : By using area of the triangle formula
 
1
3(8 4) ( 2)(4 ( 2)) 0(( 2) 8)
2
       
 
1
12 12 0
2
  
The area of the triangle is 0. Hence the three points are collinear or the lie on the same line.
Example-21. If A(5, 7), B(4,5), C(1, 6) and D(4,5) are the vertices of a quadrilateral.
Then, find the area of the quadrilateral ABCD.
Solution : By joining B to D, you willl get two triangles ABD, and BCD.
The area of ABD
 
1
5( 5 5) ( 4)(5 7) 4(7 5)
2
        
1 106
(50 8 48) 53
2 2
     square units
A
CB
D
C

184. Mathematic1 8 0
Free Distribution by A.P. Government
Also, The area of BCD  
1
4( 6 5) 1(5 5) 4( 5 6)
2
        
 
1
44 10 4 19
2
    Square units
Area of ABD + area of BCD
So, the area of quadrilateral ABCD = 53+19 = 72 square units.
7.8.2. AREA OF A TRIANGLE- ‘HERON’S FORM ULA’
We know the formula for area of the triangle is
1
base height
2
  .
Any given triangle is may be a right angle triangle, equillateral triangle and isosceles triangle.
Can we calculate the area of the triangle ?
If we know the base and height directly we apply
the above formula to find the area of a triangle.
Suppose the given triangle is a scalan triangle. That
we know the lengths of the sides only and not the height.
Then, how can we find its area?
If we want to apply the above formula for that we
will have calculate its height.
For this Heron, the Ancient Greek mathematician, derived a formuls for a triangle whose
lengths of sides are a, b and c.
A S(S )(S )(S )a b c    , where S
2
a b c 

For example, we find the area of the triangle whose lengths of sides are 12m, 9m, 15m by
using Heron’s formula we get
A S(S )(S )(S )a b c    , where S
2
a b c 

S
12 9 15 36
18
2 2
m
 
  
ThenS  a = 18  12 = 6m
S  b = 18  9 = 9m
S  c = 18  15 = 3m
A 18(6)(9)(3) 2916 54   square meter..

185. Coordinate Geometry1 8 1
Free Distribution by A.P. Government
DO THIS
Find the area of the triangle whose lengths of sides are 15m, 17m, 21m (use Heron’s
Formula) and verify your answer by using the formula A =
1
2
bh.
TRY THIS
There is a slide
in a park. One of its
side walls has been
painted in some
colour. If the sides of
the wall are 15m, 11m
and 6m. Find the area
painted in colour.
EXERCISE - 7.3
1.Find the area of the triangle whose vertices are
(i)(2, 3) (1, 0), (2, 4) (ii)(5, 1), (3, 5), (5, 2)
(iii)(0, 0), (3, 0) and (0, 2)
2.Find the value of ‘K’ for which the points are collinear.
(i)(7, 2) (5, 1) (3, K) (ii)(8, 1), (K, 4), (2, 5)
(iii)(K, K) (2, 3) and (4, 1).
3.Find the area of the triangle formed by joining the mid-points of the sides of the triangle
whose vertices are (0, 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the
given triangle.
4.Find the area of the quadrilateral whose vertices, taken inorder, are (4, 2), (3, 5),
(3, 2) and (2, 3).
5.The sides of a triangular plot are in the ratio of 3:5:7 and its perimeter is 300m. Find its
area. (use Heron’s formula)
7.9 STRAIGHT LINES
Bharadwaj and Meena discussing to find solutions for a linear equation in two variable.
Bharadwaj : Can you find solutions for 2x + 3y = 12

186. Mathematic1 8 2
Free Distribution by A.P. Government
Meena : Yes, I have done this, seex 0 3 6 3
y 4 2 0 6
2x+3y = 12
3y = 12 - 2x
y =
12 2
3
x
Bharadwaj : So the solutions of this equations
are x = 0, 3, 6, 3
and y = 4, 2, 0, 6
Meena : Can you write these solutions
in order pairs
Bharadwaj : Yes, (0, 4), (3, 2), (6, 0),
(3, 6)
Meena, can you plot these points
on the coordinate plane.
Meena : I have done case like this case
Bharadwaj : What do you observe?
What this line represents?
Meena : It is a straight line.
Bharadwaj : Can you identify some more points on this line?
Can you help meena to find some points on this line?
..................., ..................., ..................., ...................
And In this line, what is AB called ?
AB is a line segment.
DO THIS
Plot these points on the coordinates axis and join Them:
(1)A(1, 2), B(3, 4), C(7, 1)
(2)P(3, 5) Q(5, 1), R(2, 1), S(1, 2)
Which gives a straight line? Which is not? why?
-9
Y
O
X
1
Y
1
X
-8-7-6-5-4-3-2-1 123456789
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9
A(0,4)
B(3,2)
C(6,0)
D(-3,6)

187. Coordinate Geometry1 8 3
Free Distribution by A.P. Government
TRY THIS
Is y = x + 7 represent a straight line? draw and show on the coordinate axes.
7.9.1 SLOPE OF THE STRAIGHT LINE
You might have seen slider in a park. Two sliders have been given here. On which slider
Can you slide faster?
Obviousely your answer will be second “Why”?
Observe these lines.
Which line makes more angle with OX ?
Since the line “m” makes more angle than line ‘l’.
line ‘m’ has more “slope” than line ‘l’. We may also be termed “Steepness” of a line as
slope.
How we find the slope of a line?
ACTIVITY
Consider the line given in the figure indentify the points on the line and fill the table below.
x coordinate1 - - 4 -
y coordinate2 3 4 - 6
l
X
O
O
X
m

188. Mathematic1 8 4
Free Distribution by A.P. Government
We can observe that coordinates
are change when x coordinates change.
When y coordinate increases from
y
1
= 2 to y
2
= 3,
So the change in y is = ........................
Then corresponding change in ‘x’ is = ...

change in
change in
y
x
 ..................
When y coordinate increases from
y
1
= 2, y
3
= 4
So, the change in y is = ..................
The corresponding change in x is ............
So,
change in
change in
y
x
 ...............
Then can you try other points on the line choose any two points and fill in the table.
y valueChange in y x Change in x
change in
change in
y
x
2 4 - 1 2 1 -
- - - - - - -
- - - - - - -
What can you conclude from above activity?
Therefore, there is a relation between the ratio of change in y to change in x on a line has
relation with angle made by it with X-axis.
Recall the concept of tan  from trigonametry you have studied
i.e.,
Oppositesideof angle
tan
adjecentsideof angle

 

7.9.2 SLOP BETWEEN TWO POINTS ON A LINE
Let A(x
1
, y
1
) and B(x
2
, y
2
) on line ‘l’ as shown in figure
-9
Y
O
X
1
Y
1
X
-8-7-6-5-4-3-2-1 123456789
-1
-2
-3
-4
-5
-6
-7
-8
-9
1
2
3
4
5
6
7
8
9

189. Coordinate Geometry1 8 5
Free Distribution by A.P. Government
The slope of a line =
change in
change in
y
x
Slope of AB = m =
2 1
2 1


y y
x x
Slope will denote with ‘m’ and the line ‘l’
makes the angle  with X axis.
So AB line segment makes the angle with
AC also .

Oppositesideof angle
tan
adjecent wideof angle

 

=
2 1
2 1
BC
AC



y y
x x

2 1
2 1
tan
y y
m
x x

  

Hence  m =
2 1
2 1
tan

 

y y
x x
It is the formula for find slope of line segment AB which is having end points are (x
1
, y
1
),
(x
2
, y
2
).
If is angle made by the line with X-axis, then n = tan .
Example-22. The end points of a line are (2, 3), (4, 5). Find the slope of the line.
Solution : Points of a line are (2, 3), (4, 5) then slope of the line
2 1
2 1
5 3 2
1
4 2 2
 
   
 
y y
m
x x
Slope of the given line is 1.
DO THIS
Find the slope of AB

with the given end points.
1.A(4, 6) B(7, 2) 2. A(8, 4), B(4, 8)
3.A(2, 5), B(1, 7)
Y
XX
Y
C
A
(
,
)
y 1
x 1
B
(
)
x
y
2
2,
x x
2 1
-
y y
2 1
-


190. Mathematic1 8 6
Free Distribution by A.P. Government
TRY THESE
Find the slope of AB

with the points lying on
1.A(2, 1), B(2, 6) 2. A(4, 2), B(4, 2)
3.A(2, 8), B(2, 2)
4.Justify that the line AB

parallel to Y-axis when the x coordinates or equal with given
points underfefined. Why?
THINK AND DISCUSS
Find the slope AB

with the points lying on A(3, 2), (8, 2)
When the line AB

parallel to X-axis ?
Think and discuss with your friends in groups.
Justify A horizontal line has slope ‘0’, why ?
EXERCISE - 7.4
1.Find the slope of the line joining the two given points
(i)(4, 8) and (5, 2)
(ii)(0, 0) and ( 3,3)
(iii)(2a, 3b) and (a, b)
(iv)(a, 0) and (0, b)
(v)A(1.4, 3.7), B(2.4, 1.3)
(vi)A(3, 2), B(6, 2)
(vii)
1 1
A 3 , 3 , B 7, 2
2 2
   
 
   
   
(viii)A(0, 4), B(4, 0)
OPTIONAL EXERCISE
[This exercise is not meant for examination]
1.Find the distance between:-
i.(-10, 2) and (-4, -7) ii.(-0.9, -7) and (8.2., 5.2)
iii.(7.7, -4.1) and (-6.1, -2.8)

191. Coordinate Geometry1 8 7
Free Distribution by A.P. Government
2.Find the mid-point of the line joining the points A (2.5, 5.25) and B (5, -1)
3.Find the mid-point of the line joining the point A (0.5, -1.5) and B (4.5, -4.5)
4.If the points A (6, 1), B (8, 2), C (9, 4) and D (P, 3) taken in order are the vertices of a
parallelogram. Find P using distance formula.
5.Check wether (5, -2), (6, 4) and (7, -2) are the vertices of a triangle and which type of
triangle you obtain?
6.Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by
(-1, 6).
7.Determine the ratio in which the line 2x + y = 4 divides the line segment joining the points
A (2, -2) and B (3, 7)
8.Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
9.Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ABC
i.Then, find the coordinates of the points P on AD such that AP : PD = 2 : 1.
ii.Find the coordinates of points Q and R on medians BE and CF respectively such
that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
WHAT WE HAVE DISCUSSED
1.The distance between two points P(x, y) and Q(x
2
, y
2
) is   
2 2
2 1 2 1
x x y y   .
2.The distance of a point from the origin is
2 2
x y.
3.The distance between two points (x
1
, y
1
) and (x
1
, y
2
) on a line parallel to Y-axis is |y
2
- y
1
|.
4.The distance between two point (x
1
, y
1
) and (x
2
, y
2
) on a line parallel to X-axis is |x
2
- x
1
|.
5.The coordinates of the point P(x, y) which divides the line segment joining the points A (x
1
,
y
1
), and B(x
2
, y
2
) internally in the ratio m
1
: m
2
are
1 2 2 1 1 2 2 1
1 2 1 2
,
m x m x m y m y
m m m m
  
 
 
 
.
6.The mid-point of the line segment joining the points P(x
1
, y
1
) and (x
2
, y
2
) is
1 2 1 2
,
2 2
x x y y  
 
 
.
7.The point divides each median in the ratio 2 : 1 is the centroid of a triangle.

192. Mathematic1 8 8
Free Distribution by A.P. Government
8.The centroid of a triangle is the point of intersection of its medians. Hence the coordinates
of the centroid are
1 2 3 1 2 3
,
3 3
x x x y y y    
 
 
.
9.The area of the triangle formed by the points (x
1
, y
1
) (x
2
, y
2
) and (x
3
, y
3
) is the numerical
value of the expression
1
2
[x
1
(y
2
- y
3
) + x
2
(y
3
- y
1
) + x
3
(y
1
- y
2
)]
10.Area of a triangle formula ‘Heron’s Formula’
A = S S- S- S-a b c  S =
2
a b c 
(a, b, c are three sides of ABC)
11.Slope of the line m =
2 1
2 1
y y
x x



194. 8.1 INTRODUCTION
Snigdha has a tall tree in
the backyard of her house. She
wishes to find out the height of
that tree but she is not sure about
how to find it. Meanwhile, her
uncle arrives at home. Snigdha
requests her uncle to help her
with the height. He thinks for a
while and then ask her to bring a mirror. He places it on the ground at a certain distance from the
base of the tree. He then asked Snigdha to stand on the otherside of the mirror at such a position
from where she is able to see the top of the tree in that mirror.
When we draw the figure from (B) girl to the mirror (C) and mirror to the tree (DE) as
above, we observe triangles ABC and DEC. Now, what can you say about these two triangles?
Are they congruent? No, because although they have the same shape but their sizes are different.
Do you know what we call the geometrical figures which have the same shape, but are not
necessarily of the same size? They are called similar figures.
Can you guess how the heights of trees, mountains or distances of far-away, such as the
Sun objects have been found out? Do you think these have been measured directly with the help
of a measuring tape? The fact is that all these heights and distances have been found out using the
idea of indirect measurements which is based on the principle of similarity of figures.
8.2 SIMILAR FIGURES
Observe the object (car) in the previous figure.
If its breadth is kept the same and the length is doubled, it appears as in fig.(ii).
A
B C
D
E
(i) (ii) (iii)
Similar Triangles8

195. Mathematic1 9 0
Free Distribution by A.P. Government
If the length of fig.(i) is kept the same and its breadth is doubled, it appears as in fig.(iii).
Now, what can you say about fig.(ii) and (iii)? Do they resemble fig.(i)? We find that the
figure is distorted. Can you say that they are similar? No, they have same shape, yet they are not
similar.
Think what a photographer does when she prints photographs of different sizes from the
same film (negative) ? You might have heard about stamp size, passport size and post card size
photographs. She generally takes a photograph on a small size film, say 35 mm., and then
enlarges it into a bigger size, say 45 mm (or 55 mm). We observe that every line segment of the
smaller photograph is enlarged in the ratio of 35 : 45 (or 35 : 55). Further, in the two photographs
of different sizes, we can see that the angles are equal. So, the photographs are similar.
Similarly in geometry, two polygons of the same number of sides are similar if their
corresponding angles are equal and their corresponding sides are in the same ratio or proportion.
A polygon in which all sides and angles are equal is called a regular polygon.
The ratio of the corresponding sides is referred to as scale factor (or representative factor).
In real life, blue prints for the construction of a building are prepared using a suitable scale factor.
THINK, DISCUSS AND WRITE
Can you give some more examples from your daily life where scale factor is used.
All regular polygons having the same number of sides are always similar. For example, all
squares are similar, all equalateral triangles are similar etc.
Circles with same radius are congruent and those with different radii are not congruent.
But, as all circles have same shape, they are all similar.
We can say that all congruent
figures are similar but all similar figures
need not be congruent.
(i) (ii) (iii)
Similar
Squares
Similar equilateral
triangles
Similar
Circles

196. Similar Triangles1 9 1
Free Distribution by A.P. Government
To understand the similarity of figures more clearly, let us perform the following activity.
ACTIVITY
Place a table directly under a lighted bulb, fitted in the ceiling in your
classroom. Cut a polygon, say ABCD, from a plane cardboard and place
it parallel to the ground between the bulb and the table. Then, a shadow of
quadrilateral ABCD is cast on the table. Mark the outline of the shadow
as quadrilateral ABCD.
Now this quadrilateral ABCD is enlargement or magnification
of quadrilateral ABCD. Further, A lies on ray OA where ‘O’ is the bulb,
Bon OB

, Con OC

and Don OD

. Quadrilaterals ABCD and ABCDare of the same
shape but of different sizes.
A corresponds to vertex A and we denote it symbotically as A A. Similarly BB,
C C and DD.
By actually measuring angles and sides, you can verify
(i)A = A  , B = B  , C = C  , D = D  and
(ii)
AB BC CD DA
A B B C C D D A
  
       
.
This emphasises that two polygons with the same number of sides are similar if
(i)All the corresponding angles are equal and
(ii)All the corresponding sides are in the same ratio (or in proportion)
Is a square similar to a rectangle? In both the figures, corresponding angles are equal but
their corresponding sides are not in the same ratio. Hence, they are not similar. For similarity of
polygons only one of the above two conditions is not sufficient, both have to be satisfied.
THINK, DISSUSS AND WRITE
Can you say that a square and a rhombus are similar? Discuss with your friends. Write
why the conditions are or are not sufficient.
DO THESE
1.Fill in the blanks with similar / not similar.
(i)All squares are ........................
A B
CD
A’B’
C’D’

197. Mathematic1 9 2
Free Distribution by A.P. Government
(ii)All equilateral triangles are ........................
(iii)All isosceles triangles are ........................
(iv)Two polygons with same number of sides are ........................ if their corresponding
angles are equal and corresponding sides are equal.
(v)Reduced and Enlarged photographs of an object are ........................
(vi)Rhombus and squares are ........................ to each other.
2.Write the True / False for the following statements.
(i)Any two similar figures are congruent.
(ii)Any two congruent figures are similar.
(iii)Two polygons are similar if their corresponding angles are equal.
3.Give two different examples of pair of
(i)Similar figures(ii) Non similar figures
8.3 SIMILARITY OF TRIANGLES
In the example of Snigdha and her uncle, we had drawn two triangles. Those two triangles
showed the property of similarity. We know that, two triangles are similar if their
(i)Corresponding angles are equal and
(ii)Corresponding sides are in the same ratio (in proportion)
In ABC and DEC in the introduction,
A = D, B =E, ACB = DCE 
Also
DE EC DC
= = = K
AB BC AC
(scale factor)
then ABC is similar to DEC
Symbolically we write ABC ~ DEC
(Symbol ‘~’ is read as “Is similar to”)
As we have stated K is a scale factor, So
if K > 1 we get enlarged figures,
K = 1 We get congruent figures and
K < 1 gives reduced (or diminished figures)
A
B C
D
E

198. Similar Triangles1 9 3
Free Distribution by A.P. Government
Further, in triangles ABC and DEC, corresponding angles are
equal. So they are called equiangular triangles. The ratio of any two
corresponding sides in two equiangular triangles is always the same.
For proving this, Basic Proportionality theorem is used. This is also
known as Thales Theorem.
To understand Basic proportionality theorem or Thales theorem,
let us do the following activity.
ACTIVITY
Take any ruled paper and draw a triangle on that with base on
one of the lines. Several lines will cut the triangle ABC. Select any
one line among them and name the points where it meets the sides
AB and AC as P and Q.
Find the ratio of
AP
PB
and
AQ
QC
. What do you observe? The
ratios will be equal. Why ? Is it always true? Try for different lines intersecting the triangle. We
know that all the lines on a ruled paper are parallel and we observe that every time the ratios are
equal.
So in ABC, if PQ || BC then
AP
PB
=
AQ
QC
.
This is known as the result of basic proportionality theorem.
8.3.1 BASIC PROPORTIONALITY THEOREM (THALES THEOREM)
Theorem-8.1 : If a line is drawn parallel to one side of a triangle to intersect the other two sides
in distinct points, then the other two sides are divided in the same ratio.
Given : In ABC, DE || BC which intersects sides AB and AC at D and E respectively.
RTP
AD AE
=
DB EC
Construction : Join B, E and C, D and then draw
DM AC and EN  AB.
Proof :Area of ADE =
1
AD EN
2
 
Area of BDE =
1
BD EN
2
 
A
P Q
B C
A
N M
D
E
B C
Basic proportionality
theorem?

199. Mathematic1 9 4
Free Distribution by A.P. Government
So
1
AD EN
ar( ADE) AD
2
1ar( BDE) BD
BD EN
2
 

 

 
...(1)
Again Area of ADE =
1
AE DM
2
 
Area of CDE =
1
EC DM
2
 
1
AE DM
ar( ADE) AE
2
1ar( CDE) EC
EC DM
2
 

 

 
...(2)
Observe that BDE and CDE are on the same base DE and between same parallels BC
and DE.
Soar(BDE) = ar(CDE) ...(3)
From (1) (2) and (3), we have
AD AE
=
DB EC
Hence proved.
Is the converse of the above theorem also true? To examine this, let us perform another
activity.
ACTIVITY
Draw an angle XAY on your note book and on ray AX, mark points B
1
, B
2
, B
3
, B
4
and B
such that
AB
1
= B
1
B
2
= B
2
B
3
= B
3
B
4
= B
4
B = 1cm (say)
Similarly on ray AY, mark points C
1
, C
2
, C
3
, C
4
and C such that
AC
1
= C
1
C
2
= C
2
C
3
= C
3
C
4
= C
4
C = 2 cm (say)
Join B
1
, C
1
and B, C.
Observe that
1 1
1 1
AB AC 1
=
B B C C 4

and B
1
C
1
|| BC

200. Similar Triangles1 9 5
Free Distribution by A.P. Government
Similarly, joining B
2
C
2
, B
3
C
3
and B
4
C
4
, you see that
2 2
2 2
AB AC 2
=
B B C C 3
 and B
2
C
2
|| BC
3 3
3 3
AB AC 3
=
B B C C 2
 and B
3
C
3
|| BC
4 4
4 4
AB AC 4
=
B B C C 1
 and B
4
C
4
|| BC
From this we obtain the following theorem called converse of the Thales theorem
Theorem-8.2 : If a line divides any two sides of a triangle in the same ratio, then the line is
parallel to the third side.
Given : In ABC, a line DE is drawn such that
AD AE
DB EC

RTP : DE || BC
Proof : Assume that DE is not parallel to BC then draw the line
DE
1
parallel to BC
So
AD AE
DB E C



(why ?)
AE AE
EC E C

 

(why ?)
Adding 1 to both sides of the above, you can see that E and E must coincide (why ?)
TRY THESE
1.E and F are points on the sides PQ and PR respectively of PQR. For each of the
following, state whether. EF ||QR or not?
(i)PE = 3.9 cm EQ = 3 cm PF = 3.6 cm and FR = 2.4 cm
(ii)PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm.
(iii)PQ = 1.28 cm PR = 2.56 cm PE = 1.8 cm and PF = 3.6 cm
Y
C
1
A
B
1
C
2
C
3
C
C
4
B
2
B
3
B
4
B
X
A
D E
E
1
B C

201. Mathematic1 9 6
Free Distribution by A.P. Government
2.In the following figures DE || BC.
(i)Find EC (ii)Find AD
Construction : Division of a line segment (using Thales theorem)
Madhuri drew a line segment. She wants to divide it
in the ratio of 3 : 2. She measured it by using a scale and
divided it in the required ratio. Meanwhile her elder sister
came. She saw this and suggested Madhuri to divide the
line segment in the given ratio without measuring it Madhuri
was puzzled and asked her sister for help to do it. Then her
sister explained you may also do it by the following activity.
ACTIVITY
Take a sheet of paper from a lined note book.
Number the lines with the bottom line numbered ‘0’.
Take a thick carboard paper (or file card or chart
strip) and place it against the given line segment AB and
transfer its length to the card. Let A
1
and B
1
denote the
points on the file card corresponding to A and B.
Now put A
1
on the ‘O’ line of the lined paper and
rotate the card unitl point B
1
falls on the 5
th
line (3 + 2).
The point where the third line touches the file card, represent it with P
1
.
Again place this card along the given line segment and transfer this point P
1
and denote it
P
1
.
So P
1
is required point which divides the given line segment in the ratio 3:2.
Now let us learn how this construction can be done.
Given a line segment AB. We want to divide it in the ratio m : n where m and n are both
positive integers. Let us take m = 3 and n = 2.
Steps :
1.Draw a ray AX making an acute angle with AB.
1
.5

.
c
m
A
B C
D
E
3

.
c
m
A
B C
D E
7
.2

.
c
m
0
9
8
7
6
5
4
3
2
1
A
1
P
1
B
1
A
1
B
1
A B
A B
X

202. Similar Triangles1 9 7
Free Distribution by A.P. Government
2.With ‘A’ as centre and with any length draw an arc on
ray AX and label the point A
1
.
3.Using the
same compass setting
and with A
1
as centre
draw another arc and locate A
2
.
4.Like this locate 5 points (=m + n) A
1
, A
2
, A
3
, A
4
, A
5
such that AA
1
= A
1
A
2
= A
2
A
3
=
A
3
A
4
= A
4
A
5
5.Join A
5
B. Now through point A
3
(m = 3) draw a line
parallel to A
5
B (by making an angle equal to AA
5
B)
intersecting AB at C and obeserve that AC : CB = 3 : 2.
Now let us solve some examples on Thales theorem and its converse.
Example-1. In ABC, DE || BC and
AD 3
.
DB 5

AC = 5.6. Find AE.
Solution : In ABC, DE || BC
AD AE
DB EC
  (by B.P.T)
but
AD 3
DB 5
 So
AE 3
EC 5

Given AC = 5.6 and AE : EC = 3 : 5.
AE 3
AC AE 5


AE 3
5.6 AE 5
(cross multiplication)
5AE = (3 5.6)  3AE
8AE = 16.8
AE =
16.8
2.1cm.
8

Example-2. In the given figure
LM || AB
A
D E
B C
A B
X
A
1
A B
X
A
1
A
2
A
3
A
4
A
5
A B
X
2C3
A
1
A
2
A
3
A
4
A
5

203. Mathematic1 9 8
Free Distribution by A.P. Government
AL = x  3, AC = 2x, BM = x  2
and BC = 2x + 3 find the value of x
Solution : In ABC, LM || AB
AL BM
LC MC
  (by B.P.T)
3 2
2 ( 3) (2 3) ( 2)
x x
x x x x
 

    
3 2
3 5
x x
x x
 
 
(cross multiplication)
(x  3) (x + 5) = (x  2) (x + 3)
x
2
+ 2x  15 = x
2
+ x  6
x = 9
DO THESE
1.What value(s) of x will make DE || AB, in the given figure ?
AD = 8x + 9, CD = x + 3
BE = 3x + 4, CE = x.
2.In ABC, DE || BC. AD = x, DB = x  2, AE = x + 2 and EC = x  1.
Find the value of x.
Example-3.The diagonals of a quadrilateral ABCD intersect each other at point ‘O’ such that
AO
BO
=
CO
DO
.
Solution : Given : In quadrilateral ABCD
AO CO
=
BO DO
.
RTP : ABCD is a trapezium.
Construction : Through ‘O’ draw a line parallel to AB which meets DA at X.
Proof : In DAB, XO || AB (by construction)

DX
XA
=
DO
OB
(by basic proportionality theorem)
A
M
L
B
C

204. Similar Triangles1 9 9
Free Distribution by A.P. Government
AX
XD
=
BO
OD
..... (1)
again
AO
BO
=
CO
DO
(given)
AO
CO
=
BO
OD
..... (2)
From (1) and (2)
AX
XD
=
AO
CO
In ADC, XO is a line such that
AX
XD
=
AO
OC
XO || DC (by converse of basic proportionality theorem)
AB || DC
In quadrilateral ABCD, AB || DC
ABCD is a trapezium (by definition)
Hence proved.
Example-4. In trapezium ABCD, AB || DC. E and F are points on non-parallel sides AD and
BC respectively such that EF || AB. Show that
AE BF
=
ED FC
.
Solution : Let us join AC to intersect EF at G.
AB || DC and EF || AB (given)
EF || DC (Lines parallel to the same line are parallel to each other)
In ADC, EG || DC
So
AE AG
=
ED GC
(by BPT) ...(1)
Similarly, In CAB, GF || AB
CG CF
=
GA FB
(by BPT) i.e.,
AG BF
=
GC FC
...(2)
From (1) & (2)
AE BF
=
ED FC
.
D
A
O
X
C
B
A
D
G
E
B
F
C

205. Mathematic2 0 0
Free Distribution by A.P. Government
EXERCISE - 8.1
1.In PQR, ST is a line such that
PS PT
=
SQ TR
and
also PST PRQ   .
Prove that PQR is an isosceles triangle.
2.In the given figure, LM || CB and LN || CD
Prove that
AM AN
=
AB AD
3.In the given figure, DE || AC and DF || AE
Prove that
BF BE
=
FE EC
.
4.In the given figure, AB || CD || EF.
given AB = 7.5 cm DC = ycm
EF = 4.5 cm, BC = x cm.
Calculate the values of x and y.
5.Prove that a line drawn through the mid-point of one side of a triangle parallel to another
side bisects the third side (Using basic proportionality theorem).
6.Prove that a line joining the midpoints of any two sides of a triangle is parallel to the third
side. (Using converse of basic proportionality
theorem)
7.In the given figure, DE || OQ and DF || OR. Show
that EF || QR.
A
D
E
B C
F
A
D E
B C
A
L
C
N
D
M
B
P
D
O
E
Q R
F
A
D
F
B
C
E
7
.5
c
m
x cm
3 cm
4
.5
c
m
y cm

206. Similar Triangles2 0 1
Free Distribution by A.P. Government
8.A, B and C are points on OP, OQ and OR
respectively such that AB || PQ and AC||PR
Show that BC || QR. (See the adjacent figure)
9.ABCD is a trapezium in which AB||DC and its
diagonals intersect each other at point ‘O’. Show
that
AO CO
=
BO DO
.
10.Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.
THINK, DISCUSS AND WRITE
Discuss with your friends that in what way similarity of triangles is different from
similarity of other polygons?
8.4 CRITERIA FOR SIMILARITY OF TRIANGLES
We know that two triangles are similar if corresponding angles are equal and corresponding
sides are proportional. For checking the similarity of two triangles, we should check for the
equality of all angles and equality of ratios of their corresponding sides. Let us make an attempt
to arrive at certain criteria for similarity of two triangles. Let us perform the following activity.
ACTIVITY
Use a protractor and ruler to draw two non congruent triangles so that each triangle has a
40° and 60° angle. Check the figures made by you by measuring the third angles of each triangle.
It should be 80° (why?)
Measure the lengths of the sides of the triangles and compute the ratios of the lengths of the
corresponding sides.
Are the triangles similar?
This activity leads us to the following criterion for similarity of two triangles.
P
A
B
Q R
C
A
B C
60
°
40
°
P
Q R
60°40°

207. Mathematic2 0 2
Free Distribution by A.P. Government
8.4.1 AAA CRITERION FOR SIMILARITY OF TRIANGLES
Theorem-8.3 : If in two triangles, corresponding angles are equal, then their corresponding
sides are in the same ratio (or propostion) and hence the two triangles are similar.
Given : In triangles ABC and DEF,
A = D, B = E and C = F
RTP :
AB BC AC
DE EF DF
 
Construction : Locate points P and Q on DE and DF respectively.
Such that AB = DP and AC = DQ. Join PQ.
Proof : ABC DPQ   (why ?)
This gives B = P = E and PQ || EF (How ?)
DP DQ
PE QF
 
(why ?)
i.e.,
AB AC
DE DF
 (why ?)
Similarly
AB BC
DE EF
 and So
AB BC AC
.
DE EF DF
 
If two angles of a triangle are respectively equal to the two angles of another triangle, then
by angle sum property of a triangle, third angles will also be equal.
So AA similarity criterion is stated as it two angles of one triangle are respectively equal to
the two angles of anther triangle, then the two triangles are similar.
What about the converse of the above statement ?
If the sides of a triangle are respectively proportional to the sides of another triangle, is it
true that their corresponding angles are equal ?
Let us exercise it through an activity.
ACTIVITY
Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm, CA = 8 cm,
DE = 4.5 cm, EF = 9 cm and FD = 12 cm.
A
B C
D
E F
P Q
A
B C
D
E
F

208. Similar Triangles2 0 3
Free Distribution by A.P. Government
So you have
AB BC CA 2
.
DE EF FD 3
  
Now measure the angles of both the triangles. What do you observe? What can you say
about the corresponding angles? They are equal, so the triangles are similar. You can verify it for
different triangles.
From the above activity, we can give the following criterion for similarity of two triangles.
8.4.2. SSS Criterion for Similarity of Triangles
Theorem-8.4 : If in two triangles, sides of one triangle are
proportional to the sides of the other triangle, then their
corresponding angles are equal and hence the triangles are
similar.
Given : Take ABC and DEF such that
AB BC CA
( 1)
DE EF FD
  
RTP :A = D, B = E, C = F
Construction : Locate points P and Q on DE and DF respectively
Such that AB = DP and AC = DQ. Join PQ.
Proof :
DP DQ
PE QF
 and PQ || EF (How ?)
So P = E and Q = F (why ?)
DP DQ PQ
DE DF EF
  
So
DP DQ BC
DE DF EF
  (why ?)
So BC = PQ (Why ?)
ABC DPQ   (why ?)
So A = D, B = E and C = F(How ?)
We studied that for similarity of two polygons any one condition is not sufficient. But for
the similarity of triangles, there is no need for fulfillment of both the conditions as one automatically
implies the other. Now let us look for SAS similarity criterion. For this, let us perform the
following activity.
A
B C
P Q

209. Mathematic2 0 4
Free Distribution by A.P. Government
ACTIVITY
Draw two triangles ABC and DEF such that AB = 2 cm, A =50
0
AC = 4cm,
DE = 3 cm, D = 50
0
and DF = 6cm.
Observe that
AB AC 2
DE DF 3
  and A = D = 50
0
.
Now measure B, C, E, F also measure BC and EF..
Observe that B = E and C = F also
BC 2
.
EF 3

So, the two triangles are similar. Repeat the same for triangles with different measurements,
which gives the following criterion for similarity of triangles.
8.4.3 SAS CRITERION FOR SIMILARITY OF TRIANGLES
Theorem-8.5 : If one angle of a triangle is equal to one angle of the other triangle and the sides
including these angles are proportional, then the two triangles are similar.
Given : In ABC and DEF
AB AC
( 1)
DE DF
  and
A = D
RTP : ABC ~ DEF
Construction : Locate points P and Q on DE and DF respectively such that AB = DP and
AC = DQ. Join PQ.
Proof : PQ || EF and ABC DPQ   (How ?)
So A = D, B = P, C = Q
 ABC ~ DEF (why ?)
A
B C
D
E
F
A
B C
D
E F
P Q

210. Similar Triangles2 0 5
Free Distribution by A.P. Government
TRY THESE
1.Are the triangles similar? If so, name the criterion of similarity. Write the similarity relation
is symbolic form.
(i) (ii)
(iii) (iv)
(v) (vi)
(vii) (viii)
2.Explain why the triangles are similar and then find the value of x.
(i) (ii)
(iii) (iv)
(v) (vi)
KG
H
I
F P
Q
R
L
M
N
A
B C
X Y
3
2 2
3
A
B C
P J
5
32
3
1
3
Q
O
P
B
A
A
B C
40°
60°80°
P
Q R
60°
40° 80°
3

c
m
.
2 cm.
2
c
m
.
6

c
m
.5

c
m
.
4 cm.
70°
5
c
m
.
2
.5
c
m
.
6
c
m
.
1
0

c
m
.
70°
3
5
x
4.5
110°
70°
3 3
5
x
x
22
24 14
2
x
9
6
x
5
4
15 18
7.5
x
12

211. Mathematic2 0 6
Free Distribution by A.P. Government
(vii) (viii)
Construction : To construct a triangle similar to a given triangle as per given scale factor.
a)Construct a triangle similar to a given triangle ABC with its sides equal to
3
4
of corresponding
sides of ABC (scale factor
3
4
)
Steps : 1. Draw a ray BX, making an acute angle with BC on the side
opposite to vertex A.
2.Locate 4 points B
1
, B
2
, B
3
and B
4
on BX so that BB
1
= B
1
B
2
=
B
2
B
3
= B
3
B
4
.
3.Join B
4
C and draw a line from B
3
to C.
Which is parallel to B
4
C and it is intersecting BC at C.
4.Draw a line through C parallel to AB to intersect AB at A.
Which is parallel to line CA.
So A B C   is the required triangle.
Let us take some examples to illustrate the use of these criteria.
Example-5. A person 1.65m tall
casts a 1.8m shadow. At the
same instance, a lamp-posts
casts a shadow of 5.4 m. Find
the height of the lamppost.
Solution: In ABC and PQR
B = Q= 90
0
.
C = R (AC || PR, all sun’s rays are parallel at any instance)
ABC ~ PQR ( by AA similarity)
AB BC
PQ QR

Giratte
15
1.5 m.Mirror
x
1
.
6

m
.
A
x
E
B C
6 cm.
4 cm.
5 cm.
A
B C
C
1
B
1
B
2
B
3
B
4
X
A
1
1.8 m.
B
A
1.65 m.
h m.
C 5.4 m.
R
P

212. Similar Triangles2 0 7
Free Distribution by A.P. Government
1.65 1.8
PQ 5.4

1.65 5.4
PQ 4.95m
1.8

 
The height of the lamp post is 4.95m.
Example-6. A man sees the top of a tower in a mirror which is at a distance of 87.6m from the
tower. The mirror is on the ground facing upwards. The man is 0.4m away from the
mirror and the distance of his eque level from the ground is 1.5m. How tall is the tower?
Solution : In ABC & EDC
ABC EDC 90    
BCA DCE   (angle of incidence and
angle of reflection are same)
ABC ~ EDC (by AA similarity)
AB BC 1.5 0.4
ED CD h 87.6
  
1.5 87.6
h 328.5m
0.4

 
Hence, the height of the towers is 328.5m.
Example7. Gopal is quite worrying about if his neighbour can see into his living room from their
house on the top floor. He has decided to build a fence that is high enough to block the
view from their top floor window. The measurements are given in the figure.
Solution : In ABC & ACE
B =C = 90°
A = A (common angle)
ABC ~ ACE (by AA similarity)
AB BD 2 BD
AC CE 8 1.2
  
2 1.2 2.4
BD 0.3m
8 8

  
Total height of the fence required is 1.5 m. + 0.3 m. = 1.8m to block the neightbour’s view.
?
E
0.4 m.C
h
B
1.5 m.
87.6 mTower
D
Mirror
A
1.5 m.
R Q
C
B A
2m.
8 m.
P
1.5 m. 1.5 m.
D
E

213. Mathematic2 0 8
Free Distribution by A.P. Government
EXERCISE - 8.2
1.In the given figure, ADE B  
(i)Show that ABC ~ ACE
(ii)If AD = 3.8 cm, AE = 3.6cm
BE = 2.1 cm EC = 4.2 cm
find DE.
2.The perimeters of two similar triangles are 30 cm and 20 cm respectively. If one side of
the first triangle is 12 cm, determine the corresponding side of the second triangle.
3.A girl of height 90 cm is walking away from the base of a lamp post at a speed of 1.2 m/
sec. If the lamp post is 3.6m above the ground, find the length of her shadow after 4
seconds.
4.CM and RN are respectively the
medians of ABC and PQR. Prove
that
(i)AMC ~ PNR
(ii)
CM AB
RN PQ

(iii)CMB ~ RNQ
5.Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the
point ‘O’ using the criterion of similarity for two triangles, show that
OA OB
OC OD
 .
6.AB, CD, PQ are perpendicular to BD.
AB = x, CD = y amd PQ = Z
prove that
1 1 1
 
x y z
.
7.A flag pole 4m tall casts a 6 m., shadow. At the same time, a nearby building casts a
shadow of 24m. How tall is the building ?
9.CD and GH are respectively the bisectors of ACD and EGF such that D and H lie
on sides AB and FE of ABC ~FEG, show that
(i)
CD AC
GH FG
 (ii)DCB ~HGE (iii) DCA ~HGF
C
E
D
A
B
M
BA
N
Q
R
P
B
Q
z
P
A
D
C
y

214. Similar Triangles2 0 9
Free Distribution by A.P. Government
9.Construct a triangle shadow similar to the given ABC, with its sides equal to
5
3
of the
corresponding sides of the triangle ABC.
10.Construct a triangle of sides 4cm, 5 cm and 6 cm. Then, construct a triangle similar to it,
whose sides are
2
3
of the corresponding sides of the first triangle.
11.Construct an Isosceles triangle whose base is 8cm and altitude is 4 cm. Then, draw another
triangle whose sides are 1
1
2
times the corresponding sides of the isosceles triangle.
12.AS and DY are altitudes of two similar triangles ABC and DEF. Prove that AX : DY =
AB : DE.
8.5 AREAS OF SIMILAR TRIANGLES
For two similar triangles, ratio of their
corresponding sides is the same. Do you think
there is any relationship between the ratio of
their areas and the ratio of their corresponding
sides ? Let us do the following activity to
understand this.
Make a list of pairs of similar polygons
in this figure.
Find
(i)The ratio of similarity and
(ii)The ratio of areas.
You will observe that ratio of areas is
the square of the ratoio of their corresponding
sides.
Let us prove it like a theorem.
Theorem-8.6 : The ratio of the areas of
two similar triangles is equal to the square
of the ratio of their corresponding sides.
Given : ABC ~ PQR
M
B
A
N
Q R
P

215. Mathematic2 1 0
Free Distribution by A.P. Government
RTP :
2 2 2
ar( ABC) AB BC CA
ar( PQR) PQ QR RP
      
  
    
     
.
Construction : Draw AM BC and PN  QR.
Proof :
1
BC AM
ar( ABC) BC AM
2
1ar( PQR) QR PN
QR PN
2
 
 
 
 
 
...(1)
In ABM & PQN
B = Q( ABC ~ PQR)
0
M N 90   
ABM ~ PQN (by AA similarity)
AM AB
PN PQ

...(2)
Also ABC ~ PQR (given)
AB BC AC
PQ QR PR
 
...(3)
ar( ABC) AB AB
ar( PQR) PQ PQ

  

from (1), (2) and (3)

2
AB
PQ
 

 
 
.
Now by using (3), we get
2 2 2
ar( ABC) AB BC AC
ar( PQR) PQ QR PR
      
  
    
     
Hence proved.
Now let us see some examples.
Example-8.Prove that if the areas of two similar triangles are equal, then they are congruent.
Solution : ABC ~ PQR

216. Similar Triangles2 1 1
Free Distribution by A.P. Government
So
2 2 2
ar( ABC) AB BC AC
ar( PQR) PQ QR PR
      
  
    
     
But
ar( ABC)
1
ar( PQR)



( areas are equal)
2 2 2
AB BC AC
1
PQ QR PR
     
  
    
    
So AB
2
= PQ
2
BC
2
= QR
2
AC
2
= PR
2
From which we get AB = PQ
BC = QR
AC = PR
ABC  PQR (by SSS congruency)
Example-9. ABC ~ DEF and their areas are respectively 64cm
2
and 121 cm
2
.
If EF = 15.4 cm., then find BC.
Solution :
2
ar( ABC) BC
ar( DEF) EF
  

 
  
2
64 BC
121 15.4
 

 
 
8 BC 8 1.4
BC 11.2cm
11 15.4 11

    .
Example-10. Diagonals of a trapezium ABCD with AB || BC, intersect each other at the point
‘O’. If AB = 2CD, find the ratio of areas of triangles AOB and COD.
Solution : In trapezium ABCD, AB || DC also AB = 2CD.
In AOB and COD
AOB = COD (vertically opposite angles)
OAB OCD   (alternate interior angles)
A B
D C
O

217. Mathematic2 1 2
Free Distribution by A.P. Government
AOB ~ COD (by AA similarity)
2
2
ar( AOB) AB
ar( COD)DC



2
2
(2DC) 4
1(DC)
 
 ar(AOB) : ar(COD) = 4 : 1.
EXERCISE - 8.3
1.Equilateral triangles are drawn on the three sides of a right angled triangle. Show that the
area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the
other two sides.
2.Prove that the area of the equilateral triangle described on the side of a square is half the
area of the equilateral triangle described on its diagonal.
3.D, E, F are mid points of sides BC, CA, AB of ABC. Find the ratio of areas of DEF
and ABC.
4.In ABC, XY || AC and XY divides the triangle into two parts of equal area. Find the
ratio of
AX
XB
.
5.Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of
their corresponding medians.
6.ABC ~ DEF. BC = 3cm EF = 4cm and area of ABC = 54 cm
2
. Determine the area
of DEF.
7.ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm. and
BP = 3cm. (area of APQ) =
1
16
(area of ABC).
8.The areas of two similar triangles are 81cm
2
and 49 cm
2
respectively. If the attitude of the
bigger triangle is 4.5 cm. Find the corresponding attitude fo the smaller triangle.
8.6 PYTHAGORAS THEOREM
You are familar with the Pythagoras theorem, you had verified this theorem through some
activities. Now we shall prove this theorem using the concept of similarity of triangles. For this,
we make use of the following result.

218. Similar Triangles2 1 3
Free Distribution by A.P. Government
Theorem-8.7 : If a perpendicular is drawn from the vertex of the right angle of a right triangle to
the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole
triangle and to each other.
ABC is a right triangle, right angled at B. Let BD be the perpendicular to hypotenuse AC.
In ADB and ABC
A = A
And ADB ABC   (why ?)
So ADB ~ ABC (How ?) ...(1)
Similarly, BDC ~ ABC (How ?) ...(2)
So from (1) and (2), triangles on both sides of the perpendicular BD are similar to the
whole triangle ABC.
Also since ADB ~ ABC
BDC ~ ABC
So ADB ~ BDC
This leads to the following theorem.
THINK, DISCUSS AND WRITE
For right angle triangle atleast one of its measurements must be an even number. Why?
Discuss between the friends and teachers.
8.6.1 BAUDHAYAN THEOREM (PY THORGORAS THEOREM)
Theorem-8.8 : In a right triangle, the square of hypotenuse is equal to the sum of the squared of
the other two sides.
Given: ABC is a right triangle right angled at B.
RTP : AC
2
= AB
2
+ BC
2
Construction : Draw BDAC.
Proof : ADB ~ ABC
AD AB
AB AC
  (sides are proportional)
AD . AC = AB
2
...(1)
Also, BDC ~ ABC
CA
B
CA
B

219. Mathematic2 1 4
Free Distribution by A.P. Government
CD BC
BC AC
 
CD . AC = BC
2
...(2)
On adding (1) & (2)
AD . AC + CD . AC = AB
2
+ BC
2
AC (AD + CD) = AB
2
+ BC
2
AC . AC = AB
2
+ BC
2
2 2 2
AC AB BC 
The above theorem was earlier given by an ancient Indian mathematician Baudhayan (about
800 BC) in the following form.
“The diagonal of a rectangle produces by itself the same area as produced by its both sides
(i.e. length and breadth).” So sometimes, this theorem is also referred to as the Baudhayan
theorem.
What about the converse of the above theorem ?
We again prove it like a theorem, as done earlier also.
Theorem-8.9 : In a triangle if square of one side is equal to the sum of squares of the other two
sides, then the angle opposite to the first side is a right angle.
Given : In ABC,
AC
2
= AB
2
+ BC
2
RTP : B = 90
0
.
Construction : Construct a right
angled triangle PQR right
angled at Q such that PQ = AB
and QR = BC.
Proof : In PQR, PR
2
= PQ
2
+ QR
2
(Pythagores theorem as Q = 90
0
)
PR
2
= AB
2
+ BC
2
(by construction)...(1)
but AC
2
= AB
2
+ BC
2
(given)
AC = PR from (1) & (2)
Now In ABC and PQR
AB = PQ (by construction)
BC = QE (by construction)
AC = PR (proved)
Q
P
RBC
A

220. Similar Triangles2 1 5
Free Distribution by A.P. Government
ABC PQR    (by SSS congruency)
 B = Q (by cpct)
but Q = 90° (by construction)
 B= 90°.
Hence proved.
Now let us take some examples.
Example-11. A ladder 25m coins reaches a window of building 20m above the ground.
Determine the distance of the foot of the ladder from the building.
In ABC, C = 90°
2 2 2
AB AC BC   (by Pythogorous theorem)
25
2
= AC
2
+ 20
2
AC
2
= 625 400 = 225
AC 225 15m 
Hence, the foot of the ladder is at a distance of 15m from the building.
Example-12. BL and CM are medians of a triangle ABC right angled at A.
Prove that 4(BL
2
+ CM
2
) = 5BC
2
.
Solution : BL and CM are medians of ABC in which A = 90°.
From ABC
BC
2
= AB
2
+ AC
2
(Pythagores theorem)...(1)
From ABL,BL
2
= AL
2
+ AB
2
So
2
2 2AC
BL AB
2
 
 
 
 
( L is the midpoint of AC)

2
2 2AC
BL AB
4
 
 4BL
2
= AC
2
+ 4AB
2
...(2)
From CMA,CM
2
= AC
2
+ AM
2
CM
2
=
2
2AB
AC
2
 

 
 
(M is the mid point of AB)
A
CB
20 m.
25 m.

221. Mathematic2 1 6
Free Distribution by A.P. Government
CM
2
=
2
2AB
AC
4

4CM
2
= 4AC
2
+ AB
2
...(3)
On adding (2) and (3), we get
4(BL
2
+ CM
2
) = 5(AC
2
+ AB
2
)
 4(BL
2
+ CM
2
) = 5BC
2
from (1).
Example-13. ‘O’ is any point inside a rectangle ABCD.
Prove that OB
2
+ OD
2
= OA
2
+ OC
2
Solution : Through ‘O’ draw PQ || BC so that P lies on AB and Q lies on DC.
Now PQ || BC
 PQ AB & PQ  DC (B= C = 90°)
So, BPQ 90   & CQP 90  
 BPQC and APQD are both rectangles.
Now from OPB. OB
2
= BP
2
+ OP
2
...(1)
Similarly from OQC, we have OC
2
= OQ
2
+ DQ
2
...(2)
From OQC, we have OC
2
= OQ
2
+ CQ
2
...(3)
And from OAP, OA
2
= AP
2
+ OP
2
Adding (1) & (2)
OB
2
+ OD
2
= BP
2
+ OP
2
+ OQ
2
+ DQ
2
= CQ
2
+ OP
2
+ OQ
2
+ AP
2
(BP = CQ and DQ = AP)
= CQ
2
+ OQ
2
+ OP
2
+ AP
2
= OC
2
+ OA
2
(from (3) & (4))
DO THESE
1.In ACB, C = 90
0
and CD  ABAB
Prove that
2
2
BC BD
.
ADAC

2.A ladder 15m long reaches a window which is 9 m above the ground on one side of a
street. Keeping its foot at the same point, the ladder is turned to other side of the street to
reach a window 12m high. Find the width of the street.
BA
C

222. Similar Triangles2 1 7
Free Distribution by A.P. Government
3.In the given fig. if AD BC
Prove that AB
2
+ CD
2
= BD
2
+ AC
2
.
Example-14. The hypotenuse of a right triangle is 6m more than twice of the shortest side. If
the third side is 2 m., which is less than the hypotenuse, find the sides of the triangle.
Solution : Let the shortest side be x m.
Then htypotenuse = (2x + 6)m and third side = (2x + 4)m.
by Pythagores theorem, we have
(2x + 6)
2
= x
2
+ (2x + 4)
2
4x
2
+ 24x + 36 = x
2
+ 4x
2
+ 16x + 16
x
2
- 8x  20 = 0
(x  10) (x + 2) = 0
x = 10 or x = 2
but x can’t be negative as side of a triangle
 x = 10
Hence, the sides of the triangle are 10m, 26m and 24m.
Example-15. ABC is a right triangle right angled at C. Let BC = a, CA = b, AB = c and let p
be the length of perpendicular from C on AB. Prove that (i) Cp = ab (ii) 2 2 2
1 1 1
p a b
 
.
Solution :
(i)Let CD  AB then CD = p.
Area of ABC =
1
AB CD
2
 
=
1
Cp
2
.
also area of ABC =
1
BC AC
2
 

1
ab
2

D
A
C
B
a
D
p
b
A
C
B
c

223. Mathematic2 1 8
Free Distribution by A.P. Government
1 1
cp ab
2 2

cp = ab ...(1)
(ii)Since ABC is a right triangle right angled at C.
AB
2
= BC
2
+ AC
2
C
2
= a
2
+ b
2
2
2 2 
 
 
 
ab
a b
p
2 2
2 2 2
1 1 1
.

  
a b
abp a b
EXERCISE - 8.4
1.Prove that the sum of the squares of the side of a rhombus
is equal to the sum of the squares of its diagonals.
2.ABC is a right triangle right angled at B. Let D and E be
any points on AB and BC respectively. Prove that AE
2
+
CD
2
= AC
2
+ DE
2
.
3.Prove that three times the square of any side of an
equilateral triangle is equal to four times the square of the
altitude.
4.PQR is a triangle right angled at P and M is a point on QR such that PM  QR .
Show that PM
2
= QM . MR.
5.ABD is a triangle right angled at A and AC  BD
Show that (i) AB
2
= BC.BD.
(ii) AC
2
= BC.DC
(iii) AD
2
= BD.CD.
6.ABC is an isosceles triangle right angled at C. Prove that AB
2
= 2AC
2
.
7.‘O’ is any point in the interior of a triangle ABC. OD 
BC, OE  AC and OF  AB, show that
(i)OA
2
+ OB
2
+ OC
2
 OD
2
 OE
2
 OF
2
= AF
2
+
BD
2
+ CD
2
(ii)AF
2
+ BD
2
+ CE
2
= AE
2
+ CD
2
+ BF
2
.
C
E
B
D
A
B
C
A
B
D
C
F
A
E

224. Similar Triangles2 1 9
Free Distribution by A.P. Government
8.A wire attached to veritical pole of height 18m is 24m long and has a stake attached to the
other end. How far from the base of the pole should the stake be driven so that the wire
will be taut?
9.Two poles of heights 6m and 11m stand on a plane ground. If the distance between the
feet of the poles is 12m find the distance between their tops.
10.In an equilateral triangle ABC, D is a point on side
BC such that BD =
1
3
BC. Prove that 9AD
2
= 7AB
2
.
11.In ABC, D and E trisect BC.
Prove that 8AE
2
= 3AC
2
+ 5AD
2
.
12.ABC is an isosceles triangle right angled at B.
Similar triangles ACD and ABE are constructed
on sides AC and AB. Find the ratio between
the areas of ABE and ACD.
OPTIONAL EXERCISE
1.In the given figure,
QT QR
PR QS
 and 1 2  
prove that PQS ~ TQR.
2.Ravi is 1.82m tall. He wants to find the height
of a tree in his backyard. From the tree’s base
he walked 12.20 m. along the tree’s shadow
to a position where the end of his shadow
exactly overlaps the end of the tree’s shadow.
He is now 6.10m from the
end of the shadow.
How tall is the tree ?
3. The diagonal AC of a parallelogram ABCD intersects
DP at the point Q, where ‘P’ is any point on side AB.
Prove that CQ  PQ = QA  QD.
C
E
B
D
A
D
C
A
B
E
21
Q R
P
T
B
A
E
D
C
12.20 m.6.10 m.
1.82 m.
C
M
A
B
P

225. Mathematic2 2 0
Free Distribution by A.P. Government
4.ABC and AMP are two right triangles right angled at B and M respectively.
Prove that (i) ABC ~AMP
(ii)
CA BC
PA MP
 .
5.An aeroplane leeves an airport and flies due north at a speed of 1000 kmph. At the same
time another aeroplane leeves the same airport and flies due west at a speed of 1200
kmph. How far apart will the two planes be after 1
1
2
hour?
6.In a right triangle ABC right angled at C. P and Q are points on sides (AC and CB
respectively) which divide these sides in the ratio of 2 : 1.
Prove that (i) 9AQ
2
= 9AC
2
+ 4BC
2
(ii) 9BP
2
= 9BC
2
+ 4AC
2
(iii) 9(AQ
2
+ BP
2
) = 13AB
2
Negation of a statement :
We have a statement and if we add “Not” after the statement, we will get a new statement;
which is called negation of the statement.
For example take a statement “ABC is a equilateral”. If we denote it by “p”, we can
write like this.
p : Triangle ABC is equilateral and its negation will be “Triangle ABC is not equilateral”.
Negation of statement p is denoted by p; and read as negotiation of p. the statement p
negates the assertion that the statement p makes.
When we write the negation of the statements we would be careful that there should no
confusion; in understanding the statement.
Observe this example carefully
P : All irrational numbers are real numbers. We can negation of p like these ways.
i)p : All irrational numbers are not real numbers.
ii)p : It is tales that all are real numbers irrational numbers.
p Not all the irrational are real numbers.
How do we decide which negation is correct? We use the following criterion “Let p be a
statement and p its negation. Then p is false whenever p is true and p is true whenever
p is false.

226. Similar Triangles2 2 1
Free Distribution by A.P. Government
For example s :2 + 2 = 4 is True
 s:2 + 2 ‘“ 4 is False
Converse of a statement :
A simple sentence which is either true or false is called a simple statements. If we combine
two simple statements the we combine two simple statements. If we combine two simple
statements then we will get a compound statement. “If” and “then” are two words which
will connect two simple sentence.
Observe this example.
p : 2 is a even number.
q : Multiples of 2 is even numbers. Combining these two simple statements p : q we can
also say p implies q and it can be denoted by p  q.
p  q: If 2 is an even number then multiples of 2 is even number.
Proof by contradiction :
In this proof by contradiction; we assume the negation of the statement as true; which we
have to prove. In the process of proving we get contradiction somewhere. Then, we
realize that this contradiction occur because of our wrong assumption which is negation is
true. Therefore we conclude that the original statement is true.
WHAT WE HAVE DISCUSSED
1.Two figures having the same shape but not necessarily of the same size are called similar
figures.
2.All the congruent figures are similar but the converse is not true.
3.Two polygons of the same number of sides are similar
If(i)their corresponding angles are equal and
(ii)Their corresponding sides are in the same ratio (ie propostion)
For similarity of polygons either of the above two condition is not sufficient.
4.If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct
points then the other two sides are divided in the same ratio.
5.If a line divides any two sides of a triangle in the same ratio, then the line is parllel to the
third side.

227. Mathematic2 2 2
Free Distribution by A.P. Government
6.In in two triangles, corresponding angles are equal, then their corresponding sides are in
the same ratio and hence the two triangles are similar (AAA similarity)
7.If two angles of a triangle are equal to the two angles of another triangle, then third angles
of both triangles are equal by angle sum property of at triangle.
8.In in two triangles, corresponding sides are in the serve ratio, then their corresponding
angles are equal and hence the triangles are similar. (SSS similar)
9.If one angle of triangle is equal to one angle of another triangle and the sides including these
angles are in the same ratio then the triangle are similar. (SAS similarity)
10.The ratio of areas of two similar triangles is equal to the square of the ratio of their
corresponding sides.
11.If a perpendicular is drawn from the vertex of a right the triangles on both sides of the
perpendicular are similar to the whole triangle and also to each other.
12.In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the
other two sides (Pythogores Theorem).
13.In in a triangle, square of one side is equal to the sum of the squares of the other two sides,
then the angle opposite to the first side is a right angle.
Puzzle
Draw a triangle. Join the mid-point of the triangle. You get 4 triangles Again join the mid-
points of these triangles. Repeat this process. All the triangles drawn are similar triangles.
Why ? think and discuss with your friends.

229. 9.1 INTRODUCTION
We have seen two lines mostly intersect at a point or do not intersect in a plane. In some
situations they coincide with each other.
Similarly, what happens when a curve and a line is given in a plane? You know a curve may
be a parabola as you have seen in polynomials or a simple closed curve “circle” which is a
collection of all those points on a plane that are at a constant distant from a fixed point.
You might have seen circular objects rolling on a plane creating a path. For example; riding
a bicycle, wheels of train on the track etc., where it seems to be a circle and a line. Is there a
relation exist between them?
Let us see what happens, if a circle and a line were given in a plane.
9.1.1 A LINE AND A CIRCLE
You are given a circle and a line drawn on a paper. Salman argues that there can only be 3
possible ways of presenting them.
Consider a circle ‘O’ and a line PQ, the three possibilities are given in figure below:
O
P
Q(i)
P
O
Q
A
B
(ii)
O
P
Q(iii)
A
Tangents and Secants
to a Circle9
Y
X
O

230. Mathematic2 2 4
Free Distribution by A.P. Government
In Fig.(i), the line PQ and the circle have no common point. In this case PQ is a non-
intersecting line with respect to the circle.
In Fig.(ii), the line PQ intersects the circle at two points A and B. It forms a chord on the
circle AB with two common points. In this case the line PQ is a secant of the circle.
In Fig.(iii), There is only one point A, common to the line PQ and the circle. This line is
called a tangent to the circle.
You can see that there cannot be any other position of the line with respect to the circle.
We will study the existence of tangents to a circle and also study their properties and constructions.
Do yo know?
The word ‘tangent’ comes from the latin word ‘tangere’, which means to touch and was
introduced by Danish mathematician Thomas Fineke in 1583.
DO THIS
i.Draw a circle with any radius. Draw four tangents at different
points. How many tangents can you draw to this circle ?
ii.How many tangents can you draw on the circle from a point
away from it.
iii.Which of the following are tangents to the circles
9.2 TANGENTS OF A CIRCLE
We can see that tangent can be drawn at any point lying on the circle. Can you say how
many tangents can be drawn at any point on the surface of the circle.
To understand this let us consider the following activity.
ACTIVITY
Take a circular wire and attach a straight wire AB at a point P of
the circular wire, so that the system rotate about the point P in a plane.
The circular wire represents a circle and
the straight wire AB represents a line
intersects the circle at the point P.
Put the system on a table and gently rotate the wire AB about the
point P to get different positions of the straight wire (Fig. 3). The
wire intersects the circular wire at P and at one more point from
the points Q
1
, Q
2
or Q
3
etc. So while it generally intersects two
circular wire at two points one of which is P in one particular position,
O
A
P
B
L
P
Q
M
B
A
P
A
11
A
1
B
11
B
1

231. Tangents and Secants to a Circle2 2 5
Free Distribution by A.P. Government
it intersects the circle only at the point P (See position A B  of AB). This is the position of a
tangent at the point P of the circle. You can check that in all other position of AB it will intersect
the circle at P and at another point, A’ B’ is a tangent to the circle at P.
We see that there is only one tangent to the circle at point P.
Moving wire AB in either direction from this position makes it cut the circular wire in two
points. All these are therefore secants. Tangent is a special case of a secant where the two points
of intersection of a line with a circle coincide.
DO THIS
Draw a circle and a secant PQ of the circle on a paper as
shown below. Draw various lines parallel to the secant on both
sides of it.
What happens to the length of chord coming close and closer
to the centre of the circle?
What is the longest chord?
How many tangents can you draw to a circle, which are parallel to each other ?
The common point of the tangent and the circle is called the point of contact and the
tangent is said to touch the circle at the common point.
Look at the given figure of Circle and its tangents.
How many tangents can you draw to a circle at a point? How many tangents to the circle
in all? See the points of contact. Draw radii from the points of contact. Do you see anything
special about the angle between the tangents and the radii at the points of contact. All appear to
be perpendicular to the corresponding tangents. We can also prove it. Let us see how.
Theorem-9.1 : The tangent at any point of a circle is
perpendicular to the radius through the point of contact.
Given : A circle with centre O and a tangent XY to the circle
at a point P.
To prove : OP is perpendicular to XY.
Proof : Here, we will use the method that assumes that the
A
P
Q
O
Y
A
P

232. Mathematic2 2 6
Free Distribution by A.P. Government
statement is wrong and shows that such an assumption leads
to a fallacy. So we will suppose OP is not perpendicular to
XY. Take a point Q on XY other than P and join OQ.
The point Q must lie outside the circle (why?) (Note
that if Q lies inside the circle, XY becomes a secant and not a
tangent to the circle)
Therefore, OQ is longer than the radius of the circle [Why?]
i.e., OQ > OP.
This must happen for all points on the line XY. It is therefore true that OP is the shortest of
all the distances of the point O to the points of XY.
So our assumption that OP is not perpendicular to XY is false. Therefore, OP is
perpendicular to XY.
Hence proved.
Note : The line containing the radius through the point of contact is also called the ‘normal to the
circle at the point’.
TRY THIS
How can you prove the converse of the above theorem.
“If a line in the plane of a circle is perpendicular to the radius at its endpoint on the circle,
then the line is tangent to the circle”.
We can find some more results using the above theorem
(i)Since there can be only one perpendicular OP at the point P, it follows that one and only
one tangent can be drawn to a circle at a given point on the circumference.
(ii)Since there can be only one perpendicular to XY at the point P, it follows that the
perpendicular to a tangent at its point of contact passes through the centre.
Think about these. Discuss these among your friends and with your teachers.
9.2.1 CONSTRUCTION OF TANGENT TO A CIRCLE
How can we construct a line that would be tangent to a circle at a given point on it? We use
what we just found the tangent has to be perpendicular to the radius at the point of contact. To
draw a tangent through the point of contact we need to draw a line prependicular to the radius at
that point. To draw this radius we need to know the center of the circle. Let us see the steps for
this construction.
Construction : Construct a tangent to a circle at a given point when the centre of the circle is
known.
O
Y
A
QP

233. Tangents and Secants to a Circle2 2 7
Free Distribution by A.P. Government
Given : We have a circle with centre ‘O’ and a point P anywhere on its circumference. Then we
have to construct a tangent through P.
Steps of Construction :
1. Draw a circle with centre ‘O’ and
mark a point ‘P’ anywhere on it. This
is the circle with the point P on it. Join
OP.
2. Draw a perpendicular line through
the point P and name it as XY.
3. XY is the required tangent to the given circle passing through
P.
Can you draw one more tangent through P ? give reason.
TRY THIS
How can you draw the tangent to a cricle at a given point when
the centre of the circle is not known?
Hint : Draw equal angles QPX and PRQ . Explain the
construction.
9.2.2 FINDING LENGTH OF THE TANGENT
Can we find the length of the tangent to a circle from a given
point? Is the length of tangents from a given point to the circle the
same? Let us examine this.
We find the length of the tangent to a circle with centre ‘O’ and with as its radius = 6 cm.
from a point P such that OP = 10 cm. then find PA.
Solution : Tangent is perpendicular to the radius at the point of contact (Theorem 10.1)
Here PA is tangent segment and OA is radius of circle
OA PA OAP 90     
Now in OAP, OP
2
= OA
2
+ PA
2
(pythagoras theorem)
10
2
= 6
2
+ PA
2
100
2
= 36 + PA
2
PA
2
= 100 - 36
= 64
PA = 64 = 8 cm.
O
P
O P
X
Y
R
Q
P
X
Y
A
O
P
10
6

234. Mathematic2 2 8
Free Distribution by A.P. Government
EXERCISE - 9.1
1.Fill in the blanks
(i)A tangent to a circle intersects it in ................ point (s).
(ii)A line intersecting a circle in two points is called a .............
(iii)A circle can have ................ parallel tangents at the most.
(iv)The common point of a tangent to a circle and the circle is called ...............
(v)We can draw ............. tangents to a given circle.
2.A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a
point Q so that OQ = 12 cm. Find length of PQ.
3.Draw a circle and two lines parallel to a given line such that one is a tangent and the other,
a secant to the circle.
4.Calculate the length of tangent from a point 15 cm. away from the centre of a circle of
radius 9 cm.
5.Prove that the tangents to a circle at the end points of a diameter are parallel.
9 . 3NUMBER OF TANGENT ON A CIRCLE FROM ANY POINT
To get an idea of the number of tangents from a point on a circle, Let us perform the
following activity.
ACTIVITY
(i)Draw a circle on a paper. Take a point
P inside it. Can you draw a tangent to
the circle through this point ? You will
find that all the lines through this point
intersect the circle in two points. What
are these ? These are all secants of a
circle. So, it is not possible to draw
any tangent to a circle through a point
inside it. (See the adjacent figure)
(ii)Next, take a point P on the circle and
draw tangents through this point. You
have observed that there is only one
tangent to the circle at a such a point.
(See the adjaced figure)
P
(i)
P
O
(ii)

235. Tangents and Secants to a Circle2 2 9
Free Distribution by A.P. Government
(iii)Now, take a point P outside the circle
and try to draw tangents to the circle
from this point. What do you observe?
You will find that you can draw exactly
two tangents to the circle through this
point (See the adjacent figure)
Now, we can summarise these facts as follows :
Case (i) :There is no tangent to a circle passing through a point lying inside the circle.
Case(ii) :There is one and only one tangent to a circle passing through a point lying on the
circle.
Case(iii):There are exactly two tangents to a circle through a point lying outside the circle in
this case, A and B are the points of contacts of the tangents PA and PB respectively.
The length of the segment of the tangent from the external point P and the point of contact
with the circle is called the length of the tangent from the point P to the circle.
Note that in the above figure (iii), PA and PB are the length of the tangents from P to the
circle. What is the relation between lengths PA and PB.
Theorem-9.2 : The lengths of tangents drawn from an external point to a circle are equal.
Given : A circle with centre O, P is a point lying outside the circle and PA and PB are two
tangents on the circle from P. (See figure)
To prove : PA = PB
Proof : Join OA, OB and OP.
0
OAP OBP 90    (Angle between radii and tangents
Now in two right trianglesaccording to theorem 9.1)
OAP and OBP,
OA = OB(radii of same circle)
OP = OP(Common)
Therfore, By R.H.S. Congruency axiom,
OAP  OBP
This gives PA = PB (CPCT)
Hence Proved.
TRY THIS
Use pythagoras theorem to prove the above theorem.
O
A
P
B
(iii)
A
O
B
P

236. Mathematic2 3 0
Free Distribution by A.P. Government
9.3.1. CONSTRUCTION OF TANGENTS TO A CIRLCE
You saw that if the point lies outside the circle, there will be exactly two tangents to the
circle from this point. We shall now see how to draw these tangents.
Construction : To construct the tangents to a circle from a point outside it.
Given : We are given a circle with centre ‘O’ and a point P outside it. We have to construct two
tangents from P on to the circle.
Steps of construction :
Step(i) : Join PO and draw a perpendicular
bisector of it. Let M is the midpoint of
PO.
Step (ii) : Taking M as centre and PM or MO
as radius, draw a circle. Let it intersect
the given circle at the points A and B.
Step (iii) : Join PA and PB. Then PA and PB
are the required two tangents.
Proof : Now, Let us see how this construction
is justified.
Join OA. Then PAO is an angle in
the semicircle and,
therefore, PAO = 90°.
Can we say that PA OA ?
Since, OA is a radius of the given circle,
PA has to be a tangent to the circle (By
converse theorem of 10.1)
Similarly, PB is also a tangent to the
circle.
Hence proved.
Some interesting statements about tangents and secants and their proof:
Statement-1 : The centre of a circle lies on the bisector of the angle between two tangents drawn
from a point outside it. Can you think how we can prove it?
Proof : Let PQ and PR be two tangents drawn from a point P outside of the circle with centre O
Join OQ and OR, triangles OQP and ORP are congruent because we know that,
P
M
O
P
M
O
A
B
P
M
O
A
B

237. Tangents and Secants to a Circle2 3 1
Free Distribution by A.P. Government
OQP = ORP = 90 (Theorem 9.1)
OQ = OR (Radii)
OP is common.
This means OPQ OPR   (CPCT)
Therefore, OP is the angle bisector of QPR .
Hence, the centre lies on the bisector of the angle between the two tangents.
Statement-2 : In two concentric circles, such that a chord of the bigger circle, that touches the
smaller circle is bisected at the point of contact with the smaller circle.
Can you see how is this?
Proof : We are given two concentric circles C
1
and C
2
with centre
O and a chord AB of the larger circle C
1
, touching the smaller circle
C
2
at the point P (See figure) we need to prove that AP = PB.
Join OP.
Then AB is a tangent to the circle C
2
at P and OP is its radius.
Therefore, by Theorem 9.1
OP AB
Now, OAP and OBP are congruent. Why? This means AP = PB. Therefore, OP is the
bisector of the chord AB, as the perpendicular from the centre bisects the chord.
Statement-3 : If two tangents AP and AQ are drawn to a circle with centre O from an external
point A then PAQ 2 POQ   .
Can you see?
Solution : We are given a circle with centre O, an external point A and two tangents AP and AQ
to the circle, where P, Q are the points of contact (See figure).
We need to prove that
PAQ 2 OPQ  
Let PAQ = 
Now, by Theorem 9.2,
AP = AQ, So APQ is an isoscecles triangle
Therefore,APQ + AQP + PAQ = 180° (Sum of three angles)
1
APQ AQP (180 )
2
     
Q
O
R
P
C
1C
2
A
O
P B
A
Q
P
O

238. Mathematic2 3 2
Free Distribution by A.P. Government
=
1
90
2
 
Also, by Theorem 9.1,
OPA 90  
So, OPQ OPA PAQ    
1 1 1
90 90 PAQ
2 2 2
 
       
 
 
This gives
1
OPQ PAQ
2
   .
Statement-4 : If a circle touches all the four sides of a quadrilateral ABCD at points PQRS.
Then AB+CD = BC + DA
Can you think how do we proceed? AB, CD, BC, DA are all
chords to a circle.
For the circle to touch all the four sides of the quadrilateral at
points PQRS, it has to be inside the quadrilateral. (See figure)
How do we proceed further?
Proof : The circle touched the sides AB, BC, CD and DA of
Quadrilateral ABCD at the points P, Q, R and S respectively as shown
Since by theorem 9.2, the two tangents to a circle drawn from a point outside it, are equal,
AP = AS
BP = BQ
DR = DS
andCR = CQ
On adding, We get
AP + BP + DR + CR = AS + BR + DS + CQ
or (AP + PB) + (CR + DR) = (BR + QC) + (DS + SA)
or AB + CD = BC + DA.
Let us do one example of analysing a situation and seeing how we would construct something.
Example-1. Draw a pair of tangents to a circle of radius 5cm which are inclined to each other
at an angle 60°.
Solution : To draw the circle and the two tangents we need to see how we proceed. We only
A B
CD
P
QS
R

239. Tangents and Secants to a Circle2 3 3
Free Distribution by A.P. Government
have the radius of the circle and the angle between the tangents. We do not know the distance of
the point from where the tangents are drawn to the circle and we do not know the length of the
tangents either. We only know the angle between the tangents. Using this, we need to find out the
distance of the point outside the circle from which we have to draw the tangents.
To begin let us consider a circle with centre
‘O’ and radius 5cm, let PA and PB are two tangents
draw from a point ‘P’ outside the circle and the
angle between them is 60°. [This is not drawn
with any specific measurements and is only
indicate.]
In this, APB = 60°.
Join OP.
As we know, OP is the Bisector of APB
60
OPA OPB 30
2

       OAP OBP  
Now In OAP,
0opp.side OA
Sin30
Hyp OP
 
1 5
2 OP
 (From trigonometric ratio)
OP = 10 cm.
Now we can draw a circle of radius 5 cm. with centre
‘O’. We then mark a point P at a distance of 10 cm.
from the center of the circle. Join OP and complete the
construction as given in construction 9.2.
PA and PB are the required pair of tangents to the given
circle.
EXERCISE - 9.2
1.Choose the correct answer and give justification for each.
(i)The angle between a tangent to a circle and the radius drawn at the point of contact is
(a)60° (b)30° (c)45° (d)90°
A
O
B
P
5 cm.
5 cm.
60°
A
O
B
P
10 cm.
M

240. Mathematic2 3 4
Free Distribution by A.P. Government
(ii)From a point Q, the length of the tangent to a circle is 24 cm. and the distance of Q from
the centre is 25 cm. The radius of the circle is
(a)7cm (b)12 cm (c)15cm (d)24.5cm
(iii)If AP and AQ are the two tangents a circle with centre O so that
POQ 110 ,   then PAQ is equal to
(a)60° (b)70°
(c)80° (d)90°
(iv)If tangents PA and PB from a point P to a circle with centre O are
inclined to each other at angle of 80°, then POA is equal to
(a)50° (b)60° (c)70° (d)80°
(v)In the figure XY and X
1
Y
1
are two parallel tangents to a circle with centre O and another
tangent AB with point of contact C intersecting XY
at A and X
1
Y
1
at B then AOB 
(a)80° (b)100°
(c)90° (d)60°
2.Two concentric circles are radii 5 cm and 3cm are
drawn. Find the length of the chord of the larger
circle which touches the smaller circle.
3.Prove that the parallelogram circumscribing a circle
is a rhombus.
4.A triangle ABC is drawn to circumscribe a circle of radius 3 cm. such
that the segments BD and DC into which BC is divided by the point
of contact D are of length 9 cm. and 3 cm. respectively (See
adjacent figure). Find the sides AB and AC.
5.Draw a circle of radius 6cm. From a point 10 cm away
from its centre, construct the pair of tangents to
the circle and measure their lengths.
6.Construct a tangent to a circle of radius 4cm
from a point on the concentric circle of radius
6cm and measure its length. Also verify the measurement by actual calculation.
7.Draw a circle with the help of a bangle, Take a point outside the circle. Construct the pair
of tangents from this point to the circle measure them. Write conclusion.
A
110°
P
Q
O
O
A
BY
1
X
1
X Y
A
B
B
C
3 cm.9 cm.

241. Tangents and Secants to a Circle2 3 5
Free Distribution by A.P. Government
8.In a right triangle ABC, a circle with a side AB as diameter
is drawn to intersect the hypotenuse AC in P. Prove
that the tangent to the circle at P bisects the side BC.
9.Draw a tangent to a given circle with center O from a
point ‘R’ outside the circle. How many tangents can be
drawn to the circle from that point?
Hint : The distance of two points to the point of contact
is the same.
9.4 SEGMENT OF A CIRCLE FORMED BY A SECANT
We have seen a line and a circle. When a line meets a
circle in only one point, it is a tangent. A secant is a line which
intersects the circle at two distint points represented in the chord.
Here ‘l’ is the secant and AB is the chord.
Shankar is making a picture by sticking pink
and blue paper. He makes many pictures. One
picture he makes is of washbasin. How much
paper does he need to make this picture? This picture can be seen in two
parts. A rectangle is there, but what is the remaining part? It is the segment of
the circle. We know how to find the area of rectangle. How do we find the
area of the segment? In the following discussion we will try to find this area.
DO THIS
Shankar made the following pictures also along with washbasin.
What shapes can they be broken into that we can find area easily?
Make some more pictures and think of the shapes they can be divided into different parts.
Lets us recall how to find the area of the following geometrical figures as given in the table.
O
A
B Q
C
P
A B
l

242. Mathematic2 3 6
Free Distribution by A.P. Government
S.No. Figure Dimentions Area
1. length = l A = lb
breadth = b
2. Side = s A = s
2
3. base = b A =
1
2
bh
4. radius = r A = r
2
9 . 4 . 1 .FINDING THE AREA OF SEGMENT OF A CIRCLE
To estimate the area of segment of a circle, Swetha made the segments by drawing secants
to the circle.
As you know a segment is a region, bounded by the arc and a chord, we can see the area
that is shaded () in fig.(i) is a minor segment, semicircle in fig.(ii) and major segment in
fig.(iii).
How do we find the area of the segment? Do the following activity.
Take a circular shaped paper and fold it along with a chord less
than the diameter and shade the smaller part as shown in in the figure.
What do we call this smaller part? It is a minor segment (APB). What do
we call the unshaded portion of the circle? Obviously it is a major segment
(AQB).
You have already come across the sector and segment in earlier classes.
The portion of some unshaded part and shaded part (minor segment) is a
sector which is the combination of a triangle and a segment.
Let OAPB be a sector of a circle with centre O and radius ‘r’ as shown
in the figure. Let the angle measure of AOB be ‘x’.
A B
l
A
B
l
A
B
l
(i) (ii) (iii)
O
A B
Q
P
O
A B
P
r r
s
s
h
r
b
l

243. Tangents and Secants to a Circle2 3 7
Free Distribution by A.P. Government
You know that area of circle when the angle measure at the centre is 360° is r
2.
So, when the degree measure of the angle at the centre is 1°, then area of sector is
1
360


×r
2
.
Therefore, when the degree measure of the angle at the centre is x°, the area of sector is
360
x

×r
2
.
Now let us take the case of the area of the segment APB of a circle with centre ‘O’ and
radius ‘r’, you can see that
Area of the segment APB= Area of the sector OAPB - Area of OAB
=
360
x

× r
2
- area of OAB
TRY THIS
How can you find the area of major segment using area of minor segment?
DO THIS
1.Find the area of sector, whose radius is 7 cm. with the given angle:
i.60° ii.30° iii.72° iv.90° v.120°
2.The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand
in -10 minutes.
Now, we will see an example to find area of segment of a circle.
Example-2. Find the area of the segment AYB showing in the adjacent figure. If radius of the
circle is 21 cm and AOB = 120
0
(Use  =
22
7
and 3 = 1.732)
Solution : Area of the segment AYB
= Area of sector OAYB  Area of OAB ...(1)
Now, are of the sector OAYB =
0
0
120 22
21 21
7360
   cm
2
= 462 cm
2
For finding the area of OAB, draw OM  AB as shown in the figure:-
Note OA = OB. Therefore, by RHS congruence,AMO = BMO
O
A
B
1
2
0
°
Y
21 .cm
21 .cm

244. Mathematic2 3 8
Free Distribution by A.P. Government
So, M is the midpoint of AB and AOM = BOM =
0 01
120 60
2
 
Let, OM = x cm
So, from OMA ,
0OM
cos60 .
OA

or,
1
21 2
x

01
cos60
2
 

 
 

or,
21
2
x
So, OM =
21
2
cm
Also,
AM
OA
= sin 60°
AM 3
=
21 2
So, AM =
21 3
2
cm.
Therefore AB = 2AM =
2 21 3
2

cm. = 213m
So, Area of OAB
1
AB OM
2
 

1 21
21 3
2 2
   cm
2
.

441
3
4
 cm
2
. ...(3)
Therefore, area of the segment AYB =
441
462 3
4
 

 
 
cm
2
.
(from (1), (2) and (3)]
 
221
88 21 3 cm
4
 
= 271.047 cm
2
A
B
M
60°
60°
O
21 .cm
21 .cm

245. Tangents and Secants to a Circle2 3 9
Free Distribution by A.P. Government
Example-3.Find the area of the segments shaded in figure, if PQ = 24 cm., PR = 7 cm. and
QR is the diameter of the circle with centre O (Take =
22
7
)
Solution : Area of the segments shaded = Area of sector OQPR - Area of triangle PQR.
Since QR is diameter, QPR = 90° (Angle in a semicircle)
So, using pythagoras Theorem
In QPR, QR
2
= PQ
2
+ PR
2
= 24
2
+ 7
2
= 576 + 49
= 625
QR = 625 = 25 cm.
Then radius ofthe circle=
1
2
QR
=
1
2
(25) =
25
2
cm.
Now, area of semicircle OQPR=
1
2
r
2
=
1 22 25 25
2 7 2 2
  
= 327.38 cm
2
..... (1)
Area of right angled triangle QPR=
1
2
× PR × PQ
=
1
2
× 7 × 24
= 84 cm
2
..... (2)
From (1) and (2), we can find
Area of the shaded segments= 327.38 - 84
= 243.38 cm
2
Example-4. A round table top has six equal designs as shown in the figure. If the radius of the
table top is 14 cm., find the cost of making the designs with paint at the rate of D5 per cm
2
. (use
3 = 1.732)
Q
P
R
O
24
.
cm
7

.
c
m

246. Mathematic2 4 0
Free Distribution by A.P. Government
Solution : We know that the radius of circumscribing circle of a regular hexagon is equal to the
length of its side.
 Each side of regular hexagon = 14 cm.
Therefore, Area of six design segments = Area of circle - Area of the regular hexagon.
Now, Area of circle= r
2
=
22
7
× 14 × 14 = 616 cm
2
..... (1)
Area of regular hexagon= 6 ×
3
4
a
2
= 6 ×
3
4
× 14 × 14
= 509.2 cm
2
..... (2)
Hence, area of six designs= 616 - 509.21 (from (1), (2)
= 106.79 cm
2
.
Therefore, cost of painting the design at the rate of D5 per cm
2
= D106.79 × 5
= D533.95
EXERCISE - 9.3
1.A chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of
the corresponding: (use  = 3.14)
i.Minor segment ii. Major segment
2.A chord of a circle of radius 12 cm. subtends an angle of 120° at the centre. Find the area
of the corresponding minor segment of the circle (use  = 3.14 and 3 = 1.732)
3.A car has two wipers which do not overlap. Each wiper has a
blade of length 25 cm. sweeping through an angle of 115°.
Find the total area cleaned at each sweep of the blades. (use
22
7
)
4.Find the area of the shaded region in figure, where ABCD is a
square of side 10 cm. and semicircles are drawn with each
side of the square as diameter (use  = 3.14)
A B
CD
10 .cm
14
.
cm
r

247. Tangents and Secants to a Circle2 4 1
Free Distribution by A.P. Government
5.Find the area of the shaded region in figure, if ABCD is a
square of side 7 cm. and APD and BPC are semicircles. (use
22
7
)
6. In figure, QACB is a
quadrant of a circle with centre O
and radius 3.5 cm. If OD = 2 cm.,
find the area of the shaded region.
(use
22
7
)
7.AB and CD are respectively area of two concentric circles of radii
21 cm. and 7 cm. with centre O (See figure). If AOB = 30°, find
the area of the shaded region. (use
22
7
)
8. Calculate the area of the designed region in figure, common
between the two quadrants of the circles of radius 10 cm. each.
(use  = 3.14)
OPTIONAL EXERCISE
[This exercise is not meant for examination]
1.Prove that the angle between the two tangents drawn from an external point to a circle is
supplementary to the angle subtended by the line - segment joining the points of contact at
the centre.
2.PQ is a chord of length 8cm of a circle of radius 5cm. The
tangents at P and Q intersect at a point T (See figure). Find
the length of TP.
3.Prove that opposite sides of a quadrilateral circumscribing a
circle subtend supplementary angles at the centre of the circle.
4.Draw a line segment AB of length 8cm. Taking A as centre,
draw a circle of radius 4cm and taking B as centre, draw another
circle of radius 3cm. Construct tangents to each circle from
the centre of the other circle.
A B
CD
7 .cm
P
A
C
B
D
O
A B
30°
O
DC
7

.
c
m
2
1

.
c
m
A B
CD
10 .cm
1
0

.
c
m
P
Q
O
T
5 cm
.
8 cm.

248. Mathematic2 4 2
Free Distribution by A.P. Government
5.Let ABC be a right triangle in which AB = 6cm, BC = 8 cm and B= 90
0
. BD is the
perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents
from A to this circle.
6.Find the area of the shaded region in the figure, given in which two
circles with centres A and B touch each other at the point C. If AC
= 8 cm. and AB = 3 cm.
7. ABCD is a rectangel
with AB = 14 cm. and
BC = 7 cm. Taking DC,
BC and AD as diameters, three semicircles are
drawn as shown in the figure. Find the area of
the shaded region.
WHAT WE HAVE DISCUSSED
In this chapter, we have studied the following points.
1.The meaning of a tangent to a circle and a secant. We have used the idea of the chord of a
circle.
2.We used the ideas of different kinds of triangles particularly right angled triangles and
isosceles traingles.
3.We learn the following:
a) The tangents to a circle is prependicular to the radius through the point of contact.
b) The lengths of the two tangents from an external point to a circle are equal.
4.We learnt to do the following:
a) To construct a tangent to a circle at a given point when the centre of the circle is known.
b) To construct the pair of tangents from an external point to a circle.
c) We learnt to understand how to prove some statements about circles and tangents. In
this process learnt to use previous results and build on them logically to build new results.
5.We have learnt
Area of segment of a circle = Area of the corresponding section - Area of the corresponding
triangle.
A
B
C
D
A B
C
14 .cm

250. 10.1 INTRO DUCTION
In classes VIII and IX, we have learnt about area, surface area and volume of solid shapes.
We did many exercises to understand what they mean. We used them in real life situations and
identified what we needed and what was to be measured or estimated. For example, to find the
quantity of paint required to white wash a room, we need the surface area and not the volume. To
find the number of boxes that would contains a quantity of grain, we need the volume and not the
area.
TRY THIS
1.Consider the following situations. In each find out whether you need volume or area and
why?
i.Quantitiy of water inside a bottle.ii.Canvas needed for making a tent.
iii.Number of bags inside the lorry.iv.Gas filled in a cylinder.
v.Number of match sticks that can be put in the match box.
2.Make 5 more examples and ask your friends to choose what they need?
We see so many things of different shapes (combination of two or more) around us. Houses
stand on pillars, storage water tanks are cylindrical and are placed on cuboidal foundations, a
cricket bat has a cylindrical handle and a flat main body, etc. Think of different things around you.
Some of these are shown below:
Mensuration10

251. Mathematic2 4 4
Free Distribution by A.P. Government
Of these objects like football have shapes where we know the surface area and volume.
We can however see that other objects can be seen as combinations of the solid shapes. So,
their surface area and volume we now have to find. The table of the solid shapes, their areas and
volumes is given later.
TRY THIS
1.Break the pictures in the previous figure into solid shapes.
2.Think of 5 more things around you that can be seen as a combination of shapes. Name the
shapes that combine to make them.
Let us recall the surface areas and volumes of different solid shapes.
S.Name of FigureLateral / CurvedTotal surfaceVolume Name
No.the solid surface area area
1.Cuboid 2h(l+b) 2(lb+bh+hl) lbh l:length
b:breadth
h:height
2. Cube 4a
2
6a
2
a
3
a:side of
the cube
3. Right Perimeter of baseLateral surfacearea of base-
prism  height area+2(area ofheight
the end surface)
4.Regular 2rh 2r(r+h) r
2
h r:radius of
circular the base
Cylinder h:height
5. Right
1
2
(perimeter ofLateral surfaces
1
3
area of -
pyramid base) slant area+area ofthe base
height the base height
6. Right rl r(l+r)
1
3
r
2
h r:radius of
circular the base
cone h:height
l:slant height
7.Sphere 4r
2
4r
2 4
3
r
3
r:radius
8.Hemisphere 2r
2
3r
2 2
3
r
3
r:radius
aa
a
l
h
b

252. Mensuration2 4 5
Free Distribution by A.P. Government
Now, let us see some examples to illustrate the shapes in the table.
Example-1.The radius of a conical tent is 7 meters and its height is 10 meters. Calculate the
length of canvas used in making the tent if width of canvas is 2m.
22
Use
7
 
 
 
 
Solution :If the radius of conical tent is given (r) = 7 metres
Height = 10 m.
 So, the slant height of the cone l
2
= r
2
+ h
2
 l=
2 2
r h
= 49 100
= 149 = 12.2 m.
Now, Surface area of the tent = rl

222
7 12.2
7
m  
= 268.4 m
2
.
Area of canvas used = 268.4m
2
It is given the width of the canvas = 2m
Length of canvas used =
Area 268.4
134.2
width 2
m 
Example-2. An oil drum is in the shape of a cylinder having the following dimensions: diameter
is 2 m. and height is 7 meters. The painter charges D3 per 1m
2
to paint the drum. Find the
total charges to be paid to the painter for 10 drums ?
Solution : It is given that diameter of the (oil drum) cylinder = 2 m.
Radius of cylinder =
2
1
2 2
d
m 
Total surface area of a cylinderical drum = 2r(r + h)

22
2 1(1 7)
7
   
= 2 
22
7
 8

253. Mathematic2 4 6
Free Distribution by A.P. Government
=
352
7
m
2
. = 70.28 m
2
So, the total surface area of a drum= 70.28 m
2
Painting change per 1m
2
= D3.
Cost of painting of 10 drums= 70.28 310
= D2108.4
Example-3.A sphere, a cylinder and a cone are of the same radius and same height. Find the
ratio of their curved surface.
Solution : Let r be the common radius of a sphere, a cone and cylinder.
Height of sphere = diameter.
Then, the height of the cone = height of cylinder = height of sphere.
= 2r.
Let l be the slant height of cone =
2 2
r h
=
2 2
(2 ) 5r r r 
S
1
= Curved surface area of sphere = 4r
2
S
2
= Curved surface area of cylinder, 2rh = 2r 2r = 4r
2
S
3
= Curved surface area of cone = rl = r 5r = 5r
2
Ratio of curved surface area as


S
1
: S
2
: S
3
= 4r
2
: 4r
2
:5r
2
= 4 : 4 :5
Example-4.A company wanted to manufacture 1000 hemispherical basins from a thin steel
sheet. If the radius of hemisperical basin is 21 cm., find the required area of steel sheet to
manufacture the above hemispherical basins?
Solution : Radius of the hemisphesical basin (r) = 21 cm
Surface area of a hemispherical basin
= 2r
2

22
2 21
7
   × 21
= 2772 cm
2
.

254. Mensuration2 4 7
Free Distribution by A.P. Government
So, surface area of a hemispherical basin
= 2772 cm
2
.
Hence, the steel sheet required for one basin= 2772 cm
2
Total area of steel sheet required for 1000 basins= 2772 × 1000
= 2772000 cm
2
= 277.2 m
2
Example-5. A right circular cylinder has base radius 14cm and height 21cm.
Find:(i)Area of base or area of each end(ii) Curved surface area
(iii) Total surface area and (iv) Volume of the right circular cylinder.
Solution :Radius of the cylinder (r) = 14cm
Height of the cylinder (h) = 21cm
Now (i)Area of base(area of each end) r
2
=
22
7
(14)
2
= 616 cm
2
(ii)Curved surface area = 2rh = 2
22
7
1421 = 1848cm
2
.
(iii)Total surface area= 2Base area + curved surface area
= 2  616 + 1848 = 3080 cm
2
.
(iv)Volume of cylinder = r
2
h = Base area height
= 616  21 = 12936 cm
3
.
Example-6. Find the volume and surface area of a sphere of radius 21cm ( =
22
7
)
Solution : Radius of sphere(r) = 2.1 cm
Surface area of sphere = 4r
2
=
222 22 21 21
4 (2.1) 4
7 7 10 10
     

21386
55.44 cm
25
 
Volume of sphere =
3 34 4 22
(2.1)
3 3 7
r   

255. Mathematic2 4 8
Free Distribution by A.P. Government

4 22
2.1 2.1 2.1 38.808
3 7
      cm
3
.
Example-7. Find the volume and the total surface area of a hemisphere of radius 3.5 cm.
22
7
 
 
 
 
Solution : Radius of sphere (r) is 3.5 cm =
7
2
cm
Volume of hemisphere =
32
3
r

32 22 7 7 7 539
89.83 cm
3 7 2 2 2 6
      
Total surface area = 3r
2

22 7 7
3
7 2 2
    =
231
2
= 115.5 cm
2
EXERCISE - 10.1
1.A joker’s cap is in the form of right circular cone whose base radius is 7cm and height is 24
cm. Find the area of the sheet required to make 10 such caps.
2.A sports company was ordered to prepare 100 paper cylinders for shuttle cocks. The
required dimensions of the cylinder are 35 cm length /height and its radius is 7 cm. Find the
required area of thin paper sheet needed to make 100 cylinders?
3.Find the volume of right circular cone with radius 6 cm. and height 7cm.
4.The lateral surface area of a cylinder is equal to the curved surface area of a cone. If the
radius be the same, find the ratio of the height of the cylinder and slant height of the cone.
5.Aself help group wants to manufacture joker’s caps (conical caps) of 3cm. radius and
4 cm. height. If the available colour paper sheet is 1000 cm
2
, than how many caps can be
manufactured from that paper sheet?
6.A cylinder and cone have bases of equal radii and are of equal heights. Show that their
volumes are in the ratio of 3:1.
7.A solid iron rod has a cylinderical shape. Its height is 11 cm. and base diameter is 7cm.
Then find the total volume of 50 rods?

256. Mensuration2 4 9
Free Distribution by A.P. Government
8.A heap of rice is in the form of a cone of diameter 12 m. and height 8 m. Find its volume?
How much canvas cloth is required to cover the heap ? (Use  = 3.14)
9.The curved surface area of a cone is 4070 cm
2
and its diameter is 70 cm. What is its slant
height?
10.2 SURFACE AREA OF THE COMBINATION OF SOLIDS
We have seen solids which are made up of combination of solids. We can
observe in our real life also like wooden things, house items, medicine capsules,
bottles, oil-tankers etc., We eat ice-cream in our daily life. Can you tell how many
solid figures are there in it? It is actually made up of cone and hemisphere. We cannot
make the ice-cream with only one shape.
Lets take another example of an oil-tanker/
water-tanker. Is it a single shaped object? You
may guess that it is made up of a cylinder with
two hemisphere at it ends.
If, for some reason you wanted to find the
surface areas or volumes or capacities of such
objects, how would you do it? We cannot classify
these shapes under any of the solids you have already studied.
As we have seen, the oil-tanker was made up of a cylinder with two hemispheres stuck at
either end. It will look like the following figure:
If we consider the surface of the newly formed
object, we would be able to see only the curved
surfaces of the two hemisphere and the curved surface
of the cylinder.
TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere
Where TSA and CSA stand for ‘total surface area’ and ‘curved surface area’ respectively.
Now look at another example.
Devarsha wants to make a toy by putting together a hemisphere and a cone. Let us see the
steps that he should be going through.

257. Mathematic2 5 0
Free Distribution by A.P. Government
First, he should take a cone and hemisphere and bring their flat faces together. Here, of
course, he should take the base radius of the cone equal to the radius of the hemisphere, for the
toy is to have a smooth surface. So, the steps would be as show below:
At the end, he got a nice round-bottomed toy. Now, if he wants to find how much paint he
should be required to colour the surface of the toy, what should he know? He needs to know the
surface area of the toy, which consists of the CSA of the hemisphere and the CSA of the cone.
So, we can say that
TSA of the toy = CSA of Hemisphere + CSA of cone
TRY THIS
-Use known solid shapes and make as many objects (by combining more than two) as
possible that you come across in your daily life.
[Hint : Use clay, or balls, pipes, paper cones, boxes like cube, cuboid etc]
THINK AND DISCUSS
A sphere is inscribed in a cylinder. Is the surface of the sphere equal to
the curved surface of the cylinder? If yes, explain how?
Example-8.A right triangle, whose base and height are 15 cm. and 20 cm. is made to revolve
about its hypotenuse. Find the volume and surface area of the double cone so formed (Use
=3.14).
Solution : Let ABC be the right angled triangle such that
AB = 15cm and AC = 20 cm
Using Pythagoras theorem in ABC we have
BC
2
= AB
2
+ AC
2
Step-1 Step-2 Step-3

258. Mensuration2 5 1
Free Distribution by A.P. Government
BC
2
= 15
2
+ 20
2
BC
2
= 225 + 400 = 625
BC = 625 = 25 cm.
Let OA = x and OB = y.
In triangles ABO and ABC, we have BOA = BAC and ABO = ABC
So , by angle - angle - criterion of similarity, we have BOA ~ BAC
Therefore,
BO OA BA
BA AC BC
 

15
15 20 25
y x
 

3
15 20 5
y x
 

3
15 5
y
 and
3
20 5
x


3
15
5
y  and
3
20
5
x 
y = 9 and x = 12.
Thus, we have
OA = 12 cm and OB = 9cm
Volume of the double cone = volume of the cone CAA’ + volume of the cone BAA’
2 21 1
(OA) OC+ (OA) OB
3 3
    
2 21 1
12 16 12 9
3 3
     
1
144(16 9)
3
  
1
3.14 144 25
3
    cm
3
= 3768 cm
3
.
1 5 c m
. 2
0
c
m
.
O
1
5

c
m
.
25 cm.
A
A’
20 cm
.
B C
y
x

259. Mathematic2 5 2
Free Distribution by A.P. Government
Surface area of the doubled cone = (Curved surface area of cone CAA’)
+ (Curved surface area of cone BAA’)
= (OAAC) + (OAAB)
= (1220) + (1215) cm
2
=420 cm
2
= 420 3.14 cm
2
= 1318.8 cm
2
.
Example-9.A wooden toy rocket is in the shape of a cone mounted on a cylinder as shown in
the adjacent figure. The height of the entire rocked is 26 cm, while the height of the conical part
is 6cm. The base of the conical position has a diameter of 5cm, while the base diameter of the
cylindrical portion is 3cm. If the conical portion is to be painted orange and the cylindrical
portion is to be painted yellow, find the area of the rocket painted with each of these color (Take
 = 3.14)
Solution : Let ‘r’ be the radius of the base of the cone and its slant height be ‘l’. Further, let r
1
be the radius of cylinder and h
1
be its height
We have,
r = 2.5 cm., h = 6 cm.
r
1
= 1.5 cm. h
1
= 20 cm.
Now,
2 2
l r h 

2 2
(2.5) 6l 
6.25 36 = 42.25 = 6.5l 
Now,area to be painted orange
= Curved surface area of the cone
= rl
= 3.14 {2.5 × 6.5}
= 51.025 cm
2
Area to be painted yellow
= Curved surface area of the cylinder + Area of the base of the cylinder
= 2r
1
h
1
+
2
1
r
= r
1
(2h
1
+ r
1
)
2.5 cm.
1.5 cm.

260. Mensuration2 5 3
Free Distribution by A.P. Government
= 3.14 × 1.5 (2×20+1.5) cm
2
= 3.14  1.5  41.5 cm
2
= 4.71  41.5 cm
2
= 195.465 cm
2
.
Therefore, area to be painted yellow = 195.465 cm
2
EXERCISE - 10.2
1.A toy is in the form of a cone mounted on a hemisphere. The diameter of the base and the
height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy.
[use  = 3.14]
2.A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at
the other end. The radius of the common base is 8 cm. and the heights of the cylindrical
and conical portions are 10 cm and 6 cm respectivly. Find the total surface area of the
solid. [use  = 3.14]
3.A medicine capsule is in the shape of a cylinder with two
hemispheres stuck to each of its ends. The length of the
capsule is 14 cm. and the width is 5 mm. Find its surface
area.
4.Two cubes each of volume 64 cm
3
are joined end to end together. Find the surface area
of the resulting cuboid.
5.A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the
external diameter of the cylinder be 1.4 m. and its length be 8 m. find the cost of painting
it on the outside at rate of D20 per m
2
.
6.A sphere, a cylinder and a cone have the same radius. Find the ratio of their curved surface
areas.
7.A hemisphere is cut out from one face of a cubical wooden block such that the diameter of
the hemisphere is equal to the length of the cube. Determine the surface area of the remaining
solid.
8.A wooden article was made by scooping out a hemiphere from each end of
a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm.
and its base is of 3.5 cm, find the total surface area of the article.

261. Mathematic2 5 4
Free Distribution by A.P. Government
10.3 VOLUME OF A COMBINATION OF SOLIDS
Let us understand volume through an example.
Suresh runs an industry in a shed which is in the
shape of a cuboid surmounted by a half cylinder. The base
of the shed is of dimensions 7 m. 15 m. and the height of
the cuboidal portion is 8 m. Find the volume of air that the
shed can hold? Further suppose the machinery in the shed
occupies a total space of 300 m
3
and there are 20 workers,
each of whom occupies about 0.08 m
3
space on an
average. Then how much air is in the shed ?
The volume of air inside the shed (when there are
no people or machinery) is given by the volume of air inside
the cuboid and inside the half cylinder taken together. The
length, breadth and height of the cuboid are 15 m., 7 m.
and 8 m. respectively. Also the diameter of the half cylinder
is 7 m. and its height is 15 m.
So the required volume= volume of the cuboid +
1
2
volume of the cylinder..
31 22 7 7
15 7 8 15
2 7 2 2
m
 
       
 
 
= 1128.75m
3
.
Next, the toal space occupied by the machinery
= 300 m
3
.
And the total space occupied by the workers
= 200.08 m
3
= 1.6m
3
Therefore, the volume of the air, when there are machinery and workers
= 1128.75 (300.00 + 1.60)
= 1128.75 - 301.60 = 827.15 m
3
Note : In calculating the surface area of combination of solids, we can not add the surface areas
of the two solids because some part of the surface areas disappears in the process of joining
them. However, this will not be the case when we calculate the volume. The volume of the solid
formed by joining two basic solids will acutally be the sum of the volumes of the constituents as
we seen in the example above.

262. Mensuration2 5 5
Free Distribution by A.P. Government
TRY THIS
1.If the diameter of the cross - section of a wire is decreased by 5%, by what percentage
should the length be increased so that the volume remains the same ?
2.Surface area of a sphere and cube are equal. Then find the ratio of their volumes.
Let us see some more examples.
Example-10. A solid toy is in the form of a right circular cylinder with hemispherical shape at
one end and a cone at the other end. Their common diameter is 4.2 cm and the height of the
cylindrical and conical portions are 12cm and 7cm respectively. Find the volume of the solid toy.
22
Use
7
 
 
 
 
.
Solution : Let height of the conical portion h
1
= 7cm
The height of cylindrical portion h
2
= 12 cm
Radius (r) =
4.2
2
= 2.1 =
21
10
cm
Volume of the solid
= Volume of the Cone + Volume of the Cylinder + Volume of the Hemisphere.
=
2 2 3
1 2
1 2
3 3
r h r h r    
=
2
1 2
1 2
3 3
r h h r
 
  
 
 
2
22 21 1 2 21
7 12
7 10 3 3 10
   
      
  
   
22 441 7 12 7
7 100 3 1 5
 
    
 
 
22 441 35 180 21
7 100 15
  
  
 
 
322 441 236 27258
218.064 cm .
7 100 15 125
    
h
1
h
2
r

263. Mathematic2 5 6
Free Distribution by A.P. Government
Example-11.A cylindrical container is filled with ice-cream whose diameter is 12 cm. and
height is 15 cm. The whole ice-cream is distributed to 10 children in equal cones having
hemispherical tops. If the height of the conical portion is twice the diameter of its base, find
the diameter of the ice-cream cone.
Solution : Let the radius of the base of conical ice cream = x cm
Then, the height of the conical ice-cream
= 2 (diameter) = 2(2x) = 4x cm
Volume of ice - cream cone
= Volume of conical porton + Volume of hemispherical portion
=
1
3
r
2
h +
2
3
r
3
2 31 2
(4 )
3 3
x x x   
3 3 3
4 2 6
3 3
x x x   
 
= 2x
3
cm
3
Diameter of cylindrical container = 12 cm
Its height (h) = 15 cm
 Volume of cylindrical container = r
2
h
= (6)
2
15
= 540
Number of children to whom ice-cream is given = 10
Volumeof cylindricalcontainer
Volume of one ice creamcone
= 10
 3
540
2x


= 10
2x
3
× 10 = 540

3540
27
2 10
x 

 x
3
= 27
x cm.
x cm.

264. Mensuration2 5 7
Free Distribution by A.P. Government
 x
3
= 3
3
 x = 3
 Diameter of ice-cream cone (2x) = 2(3) = 6cm
Example-12.A solid consisting of a right circular cone standing on a hemisphere, is placed
upright in a right circular cylinder full of water and touches the bottom. Find the volume of
water left in the cylinder, given that the radius of the cylinder is 3 cm. and its height is 6cm.
The radius of the hemisphere is 2 cm. and the height of the cone is 4 cm.
22
Take
7
 
 
 
 
.
Solution : In the figure drawn here,
ABCD is a cylinder and LMN is a Hemisphere
OLM is a cone. We know that where a solid consisting of a cone and hemisphere is
immersed in the cylinder full of water, then some water equal to the volume of the solid, is
displaced.
Volume of Cylinder = r
2
h =   32  6 = 54 cm
3
Volume of Hemisphere =
3 3 32 2 16
2 cm
3 3 3
r    
Volume of Cone =
2 2 31 1 16
2 4 cm
3 3 3
r h      
Volume of cone and hemisphere=
16 16
π + π
3 3
=
32
π
3
Volume of water left in cylinder
= Volume of Cylinder - Volume of Cone and Hemisphere
= Volume of cylinder 
32
3

= 54
32
3

=
162 32 130
3 3
   

A B
CD
O
N
L M
3 3
2
22
4

265. Mathematic2 5 8
Free Distribution by A.P. Government
3130 22 2860
136.19cm
3 7 21
   
Example-13.A cylindrical pencil is sharpened to produce a perfect cone at one end with no
over all loss of its length. The diameter of the pencil is 1cm and the length of the conical
portion is 2cm. Calculate the volume of the shavings. Give your answer correct to two
places if it is in decimal
355
use
113
 
 
 
 
.
Solution : Diameter of the pencil = 1cm
So, radius of the pencil (r) = 0.5 cm
Length of the conical portion = h = 2cm
Volume of showings = Volume of cylinder of length 2 cm and base radius 0.5 cm.
 volume of the cone formed by this cylinder
=
2 2 21 2
3 3
r h r h r h    
=
2 3 32 355
(0.5) 2 cm 1.05 cm
3 113
   
EXERCISE-10.3
1.An iron pillar consists of a cylindrical portion of 2.8 m. height and 20 cm. in diameter and
a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1 cm
3
of iron weighs
7.5 g.
2.A toy is made in the form of hemisphere surmounted by a right cone whose circular base is
joined with the plane surface of the hemisphere. The radius of the base of the cone is
7 cm. and its volume is
3
2
of the hemisphere. Calculate the height of the cone and the
surface area of the toy correct to 2 places of decimal
1
Take 3
7
 
 
 
 
.
3.Find the volume of the largest right circular cone that can be cut out of a cube whose edge
is 7 cm.
1 .cm
2

.
c
m

266. Mensuration2 5 9
Free Distribution by A.P. Government
4.A cylinderical tub of radius 5cm and length 9.8 cm is
full of water. A solid in the form of right circular cone
mounted on a hemisphere is immersed into the tub.
The radius of the hemisphere is 3.5 cm and height of
cone outside the hemisphere is 5cm. Find the volume
of water left in the tub
22
Take
7
 
 
 
 
.
5.In the adjacent figure, the height of a solid cylinder is
10 cm and diameter is 7cm. Two equal conical holes
of radius 3cm and height 4 cm are cut off as shown
the figure. Find the volume of the remaining solid.
6.Spherical Marbles of diameter 1.4 cm. are dropped into a cylindrical beaker of diameter
7 cm., which contains some water. Find the number of marbles that should be dropped in
to the beaker, so that water level rises by 5.6 cm.
7.A pen stand is made of wood in the shape of cuboid with three conical depressions to hold
the pens. The dimensions of the cuboid are 15cm by 10 cm by 3.5 cm. The radius of each
of the depression is 0.5 cm and the depth is 1.4cm. Find the volume of wood in the entire
stand.
10.4 CONVERSION OF SOLID FROM ONE SHAPE TO ANOTHER
A women self help group (DWACRA) prepares candles by melting
down cuboid shape wax. In gun factories spherical bullets are made
by melting solid cube of lead, goldsmith prepares various ornaments
by melting cubiod gold biscuts. In all these cases, the shapes of
solids are converted into another shape. In this process, the volume
always remains the same.
How does this happen? If you want a candle of any special shape,
you have to give heat to the wax in metal container till it is completely
melted into liquid. Then you pour it into another container which has the special shape that you
want.
For example, lets us take a candle in the shape of solid cylinder, melt it and pour whole of
3 cm.
4 cm.
4 cm.
3 cm.

267. Mathematic2 6 0
Free Distribution by A.P. Government
the molton wax into another container shaped like a sphere. On cooling, you will obtain a candle
in the shape of sphere. The volume of the new candle will be the same as the volume of the
earlier candle. This is what we have to remember when we come across objects which are
converted from one shape to another, or when a liquid which originally filled a container of a
particular shape is poured into another container of a different shape or size as you observe in the
following figures.
THINK AND DISCUSS
Which barrel shown n the adjacent figure can
hold more water? Discuss with your friends.
To understand what has been discussed, let us
consider some examples.
Example-14. A cone of height 24cm and radius of base 6cm is made up of modelling clay. A
child reshapes it in the form of a sphere. Find the radius of the sphere.
Solution : Volume of cone =
1
6 6 24cm
3

  
If r is the radius of the sphere, then its volume is
34
3
r
Since the volume of clay in the form of the cone and the sphere remains the same, we have
34
3
r =
1
π
3
× 6 × 6 × 24
r
3
= 3  3  24 = 3 × 3 × 3 × 8
r
3
= 3
3
 2
3
r = 3  2 = 6
Therefore the radius of the sphere is 6cm.
4
1
4
1

268. Mensuration2 6 1
Free Distribution by A.P. Government
DO THIS
1.A copper rod of diameter 1 cm. and length 8 cm. is drawn into a wire of length 18m of
uniform thickness. Find the thickness of the wire.
2.Devarshe’s house has a water tank in the shape of a cylinder on the roof. This is filled by
pumping water from a sump (an under ground tank) which is in the shape of a cuboid. The
sump has dimensions 1.57 m. 1.44 m.  9.5 cm. The water tank has radius 60 cm.
and height 95 cm. Find the height of the water left in the sump after the water tank has
been completely filled with water from the sump which had been full of water. Compare
the capacity of the tank with that of the sump.( = 3.14)
Example-15. The diameter of the internal and external surfaces of a hollow hemispherical shell
are 6 cm. and 10 cm. respectively. It is melted and recast into a solid cylinder of diameter 14 cm.
Find the height of the cylinder.
Solution : Radius of Hollow hemispherical shell =
10
2
= 5 cm. = R
Internal radius of hollow hemispherical shell =
6
2
= 3 cm. = r
Volume of hollow hemispherical shell
= External volume - Internal volume
3 32 2
R
3 3
r   
3 32
(R )
3
r  
3 32
(5 3 )
3
  
2
(125 27)
3
  
3 32 196
98cm cm
3 3

   ...(1)
Since, this hollow hemispherical shell is melted and recast into a solid cylinder. So their
volumes must be equal
Diameter of cylinder = 14 cm. (Given)
So, radius of cylinder = 7 cm.
10 cm. 6 cm.
r
r

269. Mathematic2 6 2
Free Distribution by A.P. Government
Let the height of cylinder = h
 volume of cylinder = r
2
h
=   7  7  h cm
3
= 49h cm
3
...(2)
According to given condition
volume of Hollow hemispherical shell = volume of solid cylinder

196
49
3
h   [From equatiion (1) and (2)]

196 4
cm.
3 49 3
h 

Hence, height of the cylinder = 1.33 cm.
Example-16. A hemispherical bowl of internal radius 15 cm. contains a liquid. The liquid is to
be filled into cylindrical bottles of diameter 5 cm. and height 6 cm. How many bottles are
necessary to empty the bowl ?
Solution : Volume of hemisphere =
32
3
r
Internal radius of hemisphere r = 15 cm.
 volume of liquid contained in hemispherical bowl
=
3 32
(15) cm
3

= 2250  cm
3
.
This liquid is to be filled in cylindrical bottles and the height of each bottle (h) = 6 cm.
Radius of cylindrical bottle (R) =
5
2
cm.
 Volume of 1 cylindrical bottle = R
2
h

2
5
6
2
 
  
 
 

3 325 75
6cm cm
4 2
    
Number of cylindrical bottles required

270. Mensuration2 6 3
Free Distribution by A.P. Government
=
Volume of hemispherical bowl
Volume of 1 cylindrical bottle
=
2250
75
2


=
2 2250
60
75

.
Example-17. The diameter of a metallic sphere is 6cm. It is melted and drawn into a wire
having diameter of the cross section as 0.2 cm. Find the length of the wire.
Solution : We have, diameter of metallic sphere = 6cm
 Radius of metallic shpere = 3cm
Also, we have,
Diameter of cross - section of cylindrical wire = 0.2 cm.
Radius of cross section of cylinder wire = 0.1 cm.
Let the length of wire be l cm.
Since the metallic sphere is converted into a cylidrical shaped wire of length h cm.
 Volume of the metal used in wire = Volume of the sphere
2 3 4
(0.1) 3
3
h   
2
1 4
27
10 3
h
 
   
 
 
1
36
10
h   
36 100
h



cm
= 3600 cm. = 36 m.
Therefore, the length of the wire is 36 m.
6 cm.
0.2 cm.

271. Mathematic2 6 4
Free Distribution by A.P. Government
Example-18. How many spherical balls can be made out of a solid cube of lead whose edge
measures 44 cm and each ball being 4 cm. in diameter.
Solution : Side of lead cube = 44 cm.
Radius of spherical ball =
4
2
cm. = 2 cm.
Now volume of a spherical bullet =
34
3
r
=
3 34 22
2 cm
3 7
 

34 22
8 cm
3 7
  
Volume of x sperical bullet
34 22
8 cm
3 7
x   
It is clear that volume of x sperical bullets = Volume of lead cube
34 22
8 = (44)
3 7
x   
4 22
8 = 44 44 44
3 7
x     
44 44 44
=
4 22 8
x
 

 
x = 2541
Hence, total number of sperical bullets = 2541.
Example-19. A women self help group (DWACRA) is supplied a rectangular solid (cuboid
shape) of wax with diameters 66 cm., 42 cm., 21 cm., to prepare cylindrical candles each
4.2 cm. in diameter and 2.8 cm. of height. Find the number of candles.
Solution : Volume of wax in the rectangular solid = lbh
= (66 4221) cm
3
.
Radius of cylindrical candle =
4.2
2
cm. = 2.1 cm.
Height of cylindrical candle = 2.8 cm.

272. Mensuration2 6 5
Free Distribution by A.P. Government
Volume of candle = r
2
h
=
222
(2.1) 2.8
7
 
Volume of x cylindrical wax candles =
22
2.1 2.1 2.8
7
x   
 Volume of x cylindrical candles = volume of wax in rectangular shape
22
2.1 2.1 2.8 66 42 21
7
x       
66 42 21 7
22 2.1 2.1 2.8
x
x
  

   
= 1500
Hence, the number of cylindrical wax candles is 1500.
EXERCISE - 10.4
1.A metallic sphere of radius 4.2 cm. is melted and recast into the shape of a cylinder of
radius 6cm. Find the height of the cylinder.
2.Metallic spheres of radius 6 cm., 8 cm. and 10 cm. respectively are melted to form a
single solid sphere. Find the radius of the resulting sphere.
3.A 20m deep well with diameter 7 m. is dug and the earth from digging is evenly spread out
to form a platform 22 m. by 14 m. Find the height of the platform.
4.A well of diameter 14 m. is dug 15 m. deep. The earth taken out of it has been spread
evenly all around it in the shape of a circular ring of width 7 m. to form an embankment.
Find the height of the embanknent.
5.A container shaped like a right circular cylinder having diameter 12 cm. and height 15 cm.
is full of ice cream. The icecream is to be filled into cones of height 12 cm. and diameter
6 cm., having a hemispherical shape on the top. Find the number of such cones which can
be filled with ice cream.
6.How many silver coins, 1.75 cm. in diameter and thickness 2 mm., need to be melted to
form a cuboid of dimensions 5.5 cm. 10 cm.  3.5 cm.?
7.A vessel is in the form of an inverted cone. Its height is 8 cm. and the radius of its top is 5
cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of

273. Mathematic2 6 6
Free Distribution by A.P. Government
radius 0.5cm are dropped into the vessel,
1
4
of the water flows out. Find the number of
lead shots dropped into the vessel.
8.A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller
cones, each of diameter 4
2
3
cm and height 3cm. Find the number of cones so formed.
OPTIONAL EXERCISE
[This exercise is not meant for examination purpose]
1.A golf ball has diameter equal to 4.1 cm. Its surface has 150 dimples each of radius 2 mm.
Calculate total surface area which is exposed to the surroundings. (Assume that the dimples
are all hemispherical)
22
7

 

 
 
2.A cylinder of radius 12 cm. contains water to a depth of 20 cm. A spherical iron ball is
dropped in to the cylinder and thus the level of water is raised by 6.75 cm. Find the radius
of the ball.
22
7

 

 
 
3.A solid toy is in the form of a right circular cylinder with a hemispherical shape at one end
and a cone at the other end. Their common diameter is 4.2 cm. and height of the cylindrical
and concial portion are 12 cm. and 7 cm. respectively. Find the volume of the solid
toy.
22
7

 

 
 
4.Three metal cubes with edges 15 cm., 12 cm. and 9 cm. respectively are melted together
and formed into a simple cube. Find the diagonal of this cube.
5.A hemispherical bowl of internal diameter 36 cm. contains a liquid. This liquid is to be filled
in cylindrical bottles of radius 3 cm. and height 6 cm. How many bottles are required to
empty the bowl?
WHAT WE HAVE DISCUSSED.
1.The volume of the solid formed by joining two basic solids is the sum of the volumes of the
constituents.
2.In calculating the surface area of a combination of solids, we can not add the surface area
of the two constituents, because some part of the surface area disappears in the process of
joining them.

274. 1 1 . 1INTRO DUCTIO N
We have seen triangles and their properties in previous classes.
There, we observed different daily life situations where we were using
triangles. Let’s again look at some of the daily life examples.
l Electric poles are present everywhere. They are usually erected
by using a rope. The ple, rope and the ground form a triangle.
But, if the length of the rope decreases, what will be the shape
of the triangle and what will be the angle of the rope with the
ground ?
l A person is whitewashing a wall with the help of a ladder which
is kept as shown in the adjacent figure on
left. If the person wants to paint at a higher
position, what will the person do? What will be the change in angle
of the ladder with the ground ?
lIn the temple at Jainath in Adilabad district, which was built in
13
th
century, the first rays of the Sun fall at the feet of the Idol of
Saryawarayana Swami in the month of December. There is a relation
between distance of Idol from the door, height of the hole from
which Sun rays are entering and angle of sun rays in that month. Is
there any triangle forming in this context?
l In a play ground, children like
to slip on slider and slider is on
a defined angle from earth.
What will happen to the slider if
we change the angle? Will
children still be able to play on
it?



Trigonometry11

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The above examples are geometrically showing the application part of triangles in our daily
life and we can measure the heights, distances and slopes by using the properties of triangles.
These types of problems are part of ‘triginometry’ which is a branch of mathematics. Now look
at the example of a person who is white washing the wall with the help of a ladder as shown in the
previous figure. Let us observe the following conditions.
We denote the foot of ladder by A and top of ladder by C and the point of joining height
and base as B. Therefore, ABC is a right angle triangle with right angle at B. The angle between
ladder and base is said to be .
1. If the person wants to white wash at a higher point on the wall-
l What happens to the angle made by the ladder with the ground?
l What will be the change in the distance AB?
2. If the person wants to white wash at a lower point on the wall-
l What happens to the angle made the ladder with the ground?
l What will be the change in the distance AB?
We have observed in the above example of a person who was white washing
that when he wants to paint at higher or lower points, he should change the
position of ladder. So, when ‘’ is increased, the height also
increases and the base decreases. But, when  is
decreased, the height also decreases and the base
increases. Do you agree with
this statement?
Here, we have seen a
right angle triangle ABC and
have given ordinary names to
all sides and angles. Now let’s
name the sides again because
trignometric ratios of angles are
based on sides only.
11.1.1 NAMING THE SIDES IN A RIGHT TRIANGLE
Let’s take a right triangle ABC as show in the figure.
In triangle ABC, we can consider CAB as A where angle A is an acute angle. Since
AC is the longest side, it is called “hypotenuse”.
A A B
C



276. Trigonometry2 6 9
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Here you observe the position
of side BC with respect to angle A. It is
opposite to angle A and we can call it as
“opposite side of angle A”. And the
remaining side AB can be called as
“Adjacent side of angle A”
AC = Hypotenuse
BC = Opposite side of angle A
AB = Adjacent side of angle A
DO THIS
Identify “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the
given triangles.
1. For angle R 2. (i) For angle X
(ii) For angle Y
TRY THIS
Find lengths of “Hypotenuse”, “Opposite side” and
“Adjacent side” for the given angles in the given triangles.
1. For angle C
2. For angle A
What do you observe? Is there any relation between the opposite side of the angle A and
adjacent side of angle C? Like this, suppose you are erecting a pole by giving support of strong
ropes. Is there any relationship between the length of the rope and the length of the pole? Here,
we have to understand the relationship between the sides and we will study this under the section
called trigonometric ratios.
P
RQ
Z
X Y
C
AB
C
A B


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11.2 TRIGONOMETRIC RATIOS
We have seen the example problems in the beginning of the chapter which are related to
our daily life situations. Let’s know about the trigonometric ratios and how they are defined.
ACTIVITY
1. Draw a horizontal line on a paper.
2. Let the initial point be A and mark other points B,
C, D and E at a distance of 3cm, 6cm, 9cm,
15cm respectively from A.
3. Draw the perpendiculars BP,
CQ, DR and ES of lengths
4cm, 8cm, 12cm, 16cm
from the points B, C, D and E respectively.
Length ofLength ofLength ofOpposite side Adjacent side
hypotenuseopposite sideadjacent sideHypotenuse Hypotenuse
4. Then join AP, PQ, QR and RS.
5. Find lengths of AP, AQ, AR and AS.
Then find the ratios of
BP CQ DR
, ,
AP AQ AR
and
ES
AS
.
Did you get the same ratio as
4
5
?
Similarly try to find the ratios
AB AC AD
, ,
AP AQ AR
and
AE
AS
?
S
A
B C D E
P
Q
R


278. Trigonometry2 7 1
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11.2.1 DEFINING TRIGONOM ETERIC RATIOS
In the above activity, when we observe right angle triangles ABP, ACQ, ADR and AES,
A is common, B, C, D and E are right angles and P, , Q, R and S are
also equal. Hence, we can say that triangles ABP, ACQ, ADR and AES are similar triangles.
When we observe the ratio of opposite side of angle A and hypotenuse in a right angle triangle
and the ratio of similar sides in another triangle, it is found to be constant in all the above right
angle triangles ABP, ACQ, ADR and AES. And the ratios
BP CQ DR
, ,
AP AQ AR
and
ES
AS
can be
named as “sine A” or simply “sin A” in those triangles. If the value of angle A is “x” when it was
measured, then the ratio would be “sin x”.
Hence, we can conclude that the ratio of opposite side of an angle (measure of the angle)
and length of the hypotenuse is constant in all similar right angle triangles. This ratio can
be named as “sine” of that angle.
Similarly, when we observe the ratios
AB AC AD
, ,
AP AQ AR
and
AE
AS
, it is also found to be
constant. And these are the ratios of the adjacent sides of the angle A and hypotenuses in right
angle triangles ABP, ACQ, ADR and AES. So, the ratios
AB AC AD
, ,
AP AQ AR
and
AE
AS
can be
named as “cosine A” or simply “cos A” in those triangles. If the value of the angle A is “x”, then
the ratio would be “cos x”
Hence, we can also conclude that the ratio of the adjacent side of an angle (measure of the
angle) and length of the hypotenuse is constant in all similar right triangles. This ratio
can be named as “cosine” of that angle.
Similarly, the ratio of opposite side and adjacent side of an angle is constant and it can be
named as “tangent” of that angle.
LET’S DEFINE RATIOS IN A RIGHT ANGLE TRIANGLE
Consider a right angle triangle ABC having right angle at B as shown in the following figure.
Then, trigonometric ratios of the angle A in right angle triangle ABC are defined as follows :

279. Mathematic2 7 2
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sine of A = sinA =
Length of the sideoppositetoangleA BC
Lengthof hypotenuse AC

cosine of A = cos A =
Length of the sideadjacent toangleA AB
Length of hypotenuse AC

tangent of A = tan A =
Length of the sideopposite toangleA BC
Length of the side adjacent toangle A AB

DO THIS
1. Find (i) sin C (ii) cos C and
(iii) tan C in the adjacent triangle.
2. In a triangle XYZ, Y is right angle,
XZ = 17 m and YZ = 15 cm, then find
(i) sin X (ii) cos Y (iii) tan X
3. In a triangle PQR with right angle at Q, the value of angle P is x, PQ = 7 cm and
QR = 24 cm, then find sin x and cos x.
TRY THESE
In a right angle triangle ABC, right angle is at C. BC + CA = 23 cm and BC  CA =
7cm, then find sin A and tan B.
THINK AND DISCUSS
Discuss between your friends that
(i)sin x =
4
3
for some value of angle x.
(ii)The value of sin A and cos A is always less than 1. Why?
(iii)tan A is product of tan and A.
There are three more ratios defined in trigonometry which are considered as multiplicative
inverse of the above three ratios.
C
A B
C
AB

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Multiplicative inverse of “sine A” is “cosecant A”. Simply written as “cosec A”
i.e., cosec A =
1
sin A
Similarly, multiplicative inverses of “cos A” is secant A” (simply written as “sec A”) and
that of “tan A” is “cotangent A (simply written as cot A)
i.e., sec A =
1
cos A
and cot A =
1
tan A
If sin A =
Opposite sideof theangleA
Hypotenuse
,
then cosec A =
Hypotenuse
Opposite sideof theangleA
TRY THIS
What will be the side ratios for sec A and cot A?
THINK AND DISCUSS
l Is
sin A
cosA
equal to tan A ? l Is
cosA
sin A
equal to cot A ?
Let us see some examples
Example-1. If tan A =
3
4
, then find the other trigonometric ratio of angle A.
Solution : Given tan A =
3
4
And tan A =
Opposite side
Adjacent side
Therefore, opposite side : adjacent side = 3:4
For angle A, opposite = BC = 3k
Adjacent side = AB = 4k (where k is any positive number)
Now, we have in triangle ABC (by Pythagoras theorem)
C
A B

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AC
2
= AB
2
+ BC
2
= (3k)
2
+ (4k)
2
= 25k
2
AC=
2
25k
= 5k = Hypotenuse
Now, we can easily write the other ratios of trigonometry
sin A =
3 3
5 5

k
k
and cos A =
4 4
5 5

k
k
And also cosec A =
1 5
sin A 3
, sec A =
1 5
,
cosA 4
 cot A =
1 4
tan A 3
.
Example-2. If A and P are acute angles such that sin A = sin P then prove that A = =P
Solution : Given sin A = sin P
we have sin A =
BC
AB
and sin P =
QR
PQ
Then
BC QR
AB PQ

Therefore,
BC QR
AB PQ
 = k
By using Pythagoras theorem
AC =
2 2
AB BC and PR =
2 2
PQ QR
So,
AC AB BC
PR PQ QR
 
B Q  
Example-3. Consider a triangle PQR, right angled at P, in which PQ = 29 units, QR = 21 units
and PQR = , then find the values of
(i) cos
2
 + sin
2
 and (ii) cos
2
  sin
2

C
A B
Q
P R

282. Trigonometry2 7 5
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Solution : ln PQR, we have
2 2 2 2
PR PQ QR (29) (21)   
= 8(50) 400 = 20 units
sin

 =
PR 20
PQ 29

cos


QR 21
PQ 29

Now(i) cos
2
 + sin
2
=
2 2
20 21
1
29 29
   
 
   
   
(ii) cos
2
  sin
2
=
2 2
20 21 41
29 29 841
   
 
   
   
EXERCISE - 11.1
1. In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA
respectively. Then, find out sin A, cos A and tan A.
2. The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25cm and RP = 24 cm
respectively. Then find, tan Q  tan R.
3. In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and
BAC = . Then, find cos  and tan .
4. If cos A =
12
13
, then find sin A and tan A.
5. If 3 tan A = 4, then find sin A and cos A.
6. If A and X are acute angles such that cos A = cos X then show that
A = X.
7. Given cot  =
7
8
, then evaluate (i)
(1 sin )(1 sin )
(1 cos )(1 cos )
   
   
(ii)
1 sin
cos
 

8. In a right angle triangle ABC, right angle is at B, if tan A = 3 then find the value of
(i) sin A cos C + cos A sin C(ii) cos A cos C  sin A sin C
Q
P R


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1 1 . 3TRIGONOMETRIC RATIOS OF SOME SPECIFIC ANGLES
We already know about isosceles right angle triangle and right angle triangle with angle
30º, 60º and 90º.
Can we find sin 30
o
or tan 60
o
or cos 45
o
etc. with the help of these triangles?
Does sin 0
o
or cos 0
o
exist?
11.3.1 TRIG ONOM ETRI C RATIOS OF 45
O
In isosceles right angle triangle ABC where right angle is at B
A = C = 45
o
(why ?) and BC = AB (why ?)
Let’s assume the length of BC = AB = a
Then, AC
2
= AB
2
+ BC
2
(by Pythagoras theorem)
= a
2
+ a
2
= 2a
2
,
Therefore, AC = a2
Using the definitions of trigonometric ratios,
sin 45
o
=
o
Length of theoppositesidetoangle45 BC 1
Length of hypotenuse AC 2 2
a
a
  
cos 45
o
=
o
Length of theadjacentsidetoangle45 AB 1
Lengthof hypotenuse AC 2 2
a
a
  
tan 45
o
=
o
o
Length of theoppositesidetoangle45 BC 2
1
ACLengthof theadjacentsidetoangle 45 2
a
a
  
Like this you can determine the values of cosec 45
o
, sec 45
o
and cot 45
o
.
11.3.2 TRIGONOM ETRIC RATIOS OF 30
O
AND 60
O
Let us now calculate the trigonometric ratios of
30
o
and 60
o
. To calculate them, we will take an equilateral
triangle, draw a perpendicular which can divide the triangle
into two equal right angle triangles having angles 30
o
, 60
o
and 90
o
in each.
Consider an equilateral triangle ABC. Since each
C
A
B
A
B C
D

284. Trigonometry2 7 7
Free Distribution by A.P. Government
angle is 60
o
in an equilateral triangle, we have A =B =C = 60
o
and the sides of equilateral
triangle is AB = BC = CA = 2a units.
Draw the perpendicular line AD from vertex A to BC as shown in the adjacent figure.
Perpendicular AD acts as “angle bisector of angle A” and “bisector of the side BC” in the
equilateral triangle ABC.
Therefore, BAD = CAD = 30
o
.
Since point D divides the side BC into equal halves,
BD =
1 2a
BC
2 2
  a units.
Consider right angle triangle ABD in the above given figure.
We have AB = 2a and BD = a
Then AD
2
= AB
2
 BD
2
by (Pythagoras theorem)
= (2a)
2
 (a)
2
= 3a
2
.
Therefore, AD = a3
From definitions of trigonometric ratios,
sin 60
o
=
AD 3 3
AB 2 2
 
a
a
cos 60
o
=
BD 1
AB 2 2
 
a
a
So, similarly tan 60
o
= 3 (why?)
Like the above, you can also determine the reciprocals, cosec 60
o
, sec 60
o
and cot 60
o
by using the ratio concepts.
TRY THIS
Find sin 30
o
, cos30
o
, tan 30
o
, cosec 30
o
, sec30
o
and cot 30
o
by using the ratio
concepts.
11.3.3 TRIG ONOM ETRI C RATIOS OF 0
O
A ND 90
O
Till now, we have discussed trigonometric ratios of 30
o
, 45
o
and 60
o
. Now let us
determine the trigonometric ratios of angles 0
o
and 90
o
.

285. Mathematic2 7 8
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Suppose a segment AC of length r is
making an acute angle with ray AB. Height of C
from B is BC. When AC leans more on AB so
that the angle made by it decreases, then what
happens to the lengths of BC and AB ?
As the angle A decreases, the height of
C from AB ray decreases and foot B is shifted from B to B
1
and B
2
and gradually when the angle
becomes zero, height (i.e. opposite side of the angle) will also become zero (0) and adjacent
side would be equal to AC i.e. length equal to r.
Let us look at the trigonometric ratios
sin A =
BC
AC
and cos A =
AB
AC
If A = 0
o
then BC = 0 and AC = AB = r
then sin 0
o
=
0
r
= 0 and cos 0
o
=
r
r
= 1
we know that tan A =
sin A
cosA
So, tan0
o
=
o
o
sin0 0
0
1cos0
 
THINK AND DISCUSS
Discuss between your friend about the following conditions:
1. What can you say about cosec 0
o
=
1
sin0º
? It is not defined. Why ?
C
A
B
r

C
A
B
C
A
B
C
Step (i) Step (ii)

286. Trigonometry2 7 9
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2. What can you say about cot 0
o
=
1
tan 0º
. It is again not defined. Why?
3. sec 0
o
= 1. Why ?
Now let us see what happens when angle
made by AC with ray AB increases. When angle A
is increased, height of point C increases and the foot
of the perpendicular shifts from B to X and then to Y
and so on. In other words, we can say that the
height BC increase gradually, the angle on C gets
continuous increment and at one stage the angle
reaches 90
o
. At that time, point B reaches A and AC
equal to BC.
So, when the angle becomes 90
o
, base (i.e.
adjacent side of the angle) would become zero (0), the height of C from AB ray increases and it
would be equal to AC and that is the length equal to r.
Now let us see trigonometric ratios
sin A =
BC
AC
and cos A =
AB
AC
If A = 90
o
then AB = 0 and AC = BC = r
then sin 90
o
=
r
r
= 1 and cos 90
o
=
0
r
= 0
TRY THIS
Find the ratios for tan 90
o
, cosec 90
o
, sec 90
o
and cot 90
o
.
Step (i)
Step (ii) Step (iii)

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Now, let us see the values of trigonometric ratios of all the above discussed angles in the
form of a table.
Table 11.1
A 0
o
30
o
45
o
60
o
90
o
sin A 0
1
2
1
2
3
2
1
cos A 1
3
2
1
2
1
2
0
tan A 0
1
3
1 3  (not defined)
cot A  (not defined)3 1
1
3
0
sec A 1
2
3
2 2  (not defined)
cosec A  (not defined)2 2
2
3
1
THINK AND DISCUSS
What can you say about the values of sin A and cos A, as the value of angle A increases
from 0
o
to 90
o
? (observe the above table)
If A > B, then sin A > sin B. Is it true ?
If A > B, then cos A > cos B. Is it true ? Discuss.
Example-4. In ABC, right angle is at B, AB = 5 cm and ACB = 30
o
. Determine the lengths
of the sides BC and AC.
Solution : To find the length of side BC, we
will choose the trignometric ratio involving BC
and the given side AB. Since BC is the side
adjacent to angle C and AB is the side
opposite to angle C.
Therefore,
AB
BC
= tan C
C
A
B
5

c
m

288. Trigonometry2 8 1
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i.e.
5
BC
= tan 30
o
=
1
3
which gives BC = 53 cm
Now, by using the Pythagoras theorem
AC
2
= AB
2
+ BC
2
AC
2
= 5
2
+ 53
2
AC
2
= 25 + 75
AC = 100 = 10 cm
Example-5. A chord in a circle of radius 6cm is making an angle 60
o
at the centre. Find the
length of the chord.
Solution : Given the radius of the circle OA = OB = 6cm
AOB = 60
o
OC is height from ‘O’ upon AB and it is a angle bisector.
then, COB = 30
o
.
Consider COB
sin 30
o
=
BC
OB
1 BC
2 6

BC =
6
3
2
.
But, length of the chord AB = 2BC
= 2  3 = 6 cm
 Therefore, length of the chord = 6 cm
Example-6. In ABC, right angle is at Q, PQ = 3 cm and PR = 6 cm. Determine QPR and
PRQ.
O
C
A B

289. Mathematic2 8 2
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Solution : Given PQ = 3 cm and PR = 6 cm
Therefore,
PQ
PR
= sin R
or sin R =
3 1
6 2

So, PRQ = 30
o
and therefore, QRP = 60
o
(why?)
Note :If one of the sides and any other part (either an acute angle or any side) of a right angle
triangle is known, the remaining sides and angles of the triangle can be determined.
Example-7. If sin (A  B) =
1
2
, cos (A + B) =
1
2
, 0
o
< A + B < 90
o
, A > B, find A and B.
Solution : Since sin (A  B) =
1
2
, therefore, A  B = 30
o
(why?)
Also, since cos (A + B) =
1
2
, therefore, A + B = 60
o
(why?)
Solving the above equations, we get : A = 45
o
and B = 15
o
. (How?)
EXERCISE - 11.2
1. Evaluate the following.
(i)sin 45
o
+ cos 45
o
(ii)
o
o o
cos 45
sec30 cosec60
(iii)
o o o
o o o
sin 30 tan45 cosec60
cot 45 cos60 sec30
 
 
(iv)2tan
2
45
o
+ cos
2
30
o
 sin
2
60
o
(v)
2 o 2 o
2 o 2 o
sec 60 tan 60
sin 30 cos 30


2. Choose the right option and justify your choice-
(i)
o
2 o
2tan30
1 tan 45
(a) sin 60
o
(b) cos 60
o
(c)tan 60
o
(d)sin 30
o
P
RQ

290. Trigonometry2 8 3
Free Distribution by A.P. Government
(ii)
2 o
2 o
1 tan 45
1 tan 45


(a) tan 90
o
(b) 1(c)sin 45
o
(d)0
(iii)
o
2 o
2tan30
1 tan 30
(a) cos 60
o
(b) sin 60
o
(c)tan 60
o
(d)sin 30
o
3. Evaluate sin 60
o
cos 30
o
+ sin 30
o
cos 60
o
. What is the value of sin(60
o
+ 30
o
). What
can you conclude ?
4. Is it right to say cos(60
o
+ 30
o
) = cos 60
o
cos30
o
 sin 60
o
sin 30
o
.
5. In right angle triangle PQR, right angle is at Q and PQ = 6cms RPQ = 60
o
.
Determine the lengths of QR and PR.
6. In XYZ, right angle is at Y, YZ = x, and XY = 2x then determine YXZ and
YZX.
7. Is it right to say that sin (A + B) = sin A + sin B? Justify your answer.
THINK AND DISCUSS
For which value of acute angle (i)
cos cos
4
1 sin 1 sin
 
 
   
is true?
For which value of 0
o
<  < 90
o
, above equation is not defined?
11.4 TRIGONOMETRIC RATIOS OF COMPLEMENT ARY ANGLES
We already know that two angles are said to be complementary, if their sum is equal to
90
o
in a triangle. Consider a right angle triangle ABC with right angle is at B. Are there any
complementary angles in this triangle?
Since angle B is 90
o
, sum of other two angles must be
90
o
. ( Sum of angles in a triangle 180
o
)
Therefore, A + C = 90
o
. Hence A and C
are said to be complementary angles.
Let us assume that A = x, then for angle x, BC is
opposite side and AB is adjacent side.
C
A B

291. Mathematic2 8 4
Free Distribution by A.P. Government
sin x =
BC
AC
cos x =
AB
AC
tan x =
BC
AB
cosec x =
AC
BC
sec x =
AC
AB
cot x =
AB
BC
If A + C = 90
o
, then we have C = 90
o
 A
And we have that A = x, then C = 90
o
 x
Let us look at what would be “Opposite side” and “Adjacent side” of the angle (90
o

x)in the triangle ABC.
sin(90
o
 x) =
AB
AC
cos(90
o
 x) =
BC
AC
tan(90
o
 x) =
AB
BC
Cosec(90
o
 x) =
AC
AB
sec(90
o
 x) =
AC
BC
cot(90
o
 x) =
BC
AB
Now, if we compare the ratios of angles x and (90
o
 x) from the above values of
different triginometric terms.
There can be three possibilities in above figure.
sin(90
o
 x) =
AB
AC
= cos x andcos(90
o
 x) =
BC
AC
= sin x
tan(90
o
 x) =
AB
BC
= cot x andcot(90
o
 x) =
BC
AB
= tan x
cosec(90
o
 x) =
AC
AB
= sec x andsec(90
o
 x) =
AC
BC
= cosec x
THINK AND DISCUSS
Check and discuss the above relations in the case of 0º and 90º, whether they hold for
these angles or not?
So,sin (90
o
 A) = cos A cos (90
o
 A) = sin A
tan (90
o
 A) = cot A andcot (90
o
 A) = tan A
sec (90
o
 A) = cosec A cosec (90
o
 A) = sec A

292. Trigonometry2 8 5
Free Distribution by A.P. Government
Now, let us consider some examples
Example-8. Evaluate
o
o
sec35
cosec55
Solution :cosec A = sec (90
o
 A)
cosec 55
o
= sec (90
o
 55
o
)
cosec 55
o
= sec 35
o
Now
o o
o o
sec35 sec35
cosec65 sec35
 = 1
Example-9. If cos 7A = sin(A  6
o
), where 7A is an acute angle, find the value of A.
Solution : Given cos 7A = sin(A  6
o
) ...(1)
sin (90  7A) = sin (A  6
o
)
since 90  7A & A  6
o
are both acute angles,
therefore
90
o
 7A = A  6
o
8A = 96
o
which gives A = 12
o
.
Example-10. If sin A = sin B, then prove that A + B = 90º.
Solution : Given that sin A = cos B ...(1)
We know cos B = sin (90
o
 B), we can write (1) as
sin A = sin (90
o
 B)
If A, B are acute angles, then A = 90
o
 B
 A + B = 90
o
.
Example-11. Express sin 81
o
+ tan 81
o
in terms of trigonometric ratios of angles between
0
o
and 45
o
Solution : We can write sin 81
o
= cos(90
o
 81
o
) = cos 9
o
tan 81
o
= tan(90
o
 81
o
) = cot 9
o
Then, sin 81
o
+ tan 81
o
= cos 9
o
+ cot 9
o

293. Mathematic2 8 6
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Example-12. If A, B and C are interior angles of triangle ABC, then show that
B C A
sin cos
2 2


Solution : Given A, B and C are interior angles of right angle triangle ABC then
A + B + C = 180
o
.
On dividing the above equation by 2 on both sides, we get
oA B C
90
2 2

 
oB C A
90
2 2

 
On taking sin ratio on both sides
oB C A
sin sin 90
2 2
   
    
   
B C A
sin cos
2 2
 
 
 
; hence proved.
EXERCISE 11.3
1. Evaluate
(i)
o
o
tan36
cot54
(ii)cos12
o
 sin78
o
(iii) cosec 31
o
 sec 59
o
(iv)sin 15
o
sec 75
o
(vi)tan 26
o
tan64
o
2. Show that
(i)tan 48
o
tan 16
o
tan 42
o
tan 74
o
= 1
(ii)cos36
o
cos 54
o
 sin36
0
sin 54
o
= 0.
3. If tan 2A = cot(A  18
o
), where 2A is an acute angle. Find the value of A.
4. If tanA = cot B, prove that A + B = 90
o
.
5. If A, B and C are interior angles of a triangle ABC, then show that
A B C
tan cot
2 2
 

 
 
6. Express sin 75
o
+ cos 65
o
in terms of trigonometric ratios of angles between 0
o
and 45
o
.

294. Trigonometry2 8 7
Free Distribution by A.P. Government
1 1 . 5TRIGONOMETRIC IDENTITIES
We know that an identity is that mathematical equation which is true for all the values of
the variables in the equation.
For example (a + b)
2
= a
2
+ b
2
+ 2ab is an identity.
In the same way, an identity equation having trigonometric ratios of an angle is called
trigonometric idenitity. And it is true for all the values of the angles involved in it.
Here, we will derive a trigonometric identity and remaining would be based on that.
Consider a right angle triangle ABC with right angle is at B, so
From Pythagoras theorem
We have AB
2
+ BC
2
= AC
2
....(1)
Dividing each term by AC
2
, we get
2 2 2
2 2 2
AB BC AC
AC AC AC
  
i.e.,
2 2 2
AB BC AC
AC AC AC
     
 
     
     
i.e., (cos A)
2
+ (sin A)
2
= 1
Here, we generally write cos
2
A in the place of (cos A)
2
i.e., (cos A)
2
= cos
2
A (Do not write cos A
2
)
We have given an equation having a variable A(angle) and above equation is true for all
the value of A. Hence the above equation is a trigonometric identity.
Therefore, we have trigonometric idenity
cos
2
A + sin
2
A = 1
Let us look at another trigonometric idenity
From equation (1) we have
AB
2
+ BC
2
= AC
2
2 2 2
2 2 2
AB BC AC
AB AB AB
   (Dividing each term by AB
2
)
C
A
B
B
A
B

295. Mathematic2 8 8
Free Distribution by A.P. Government

2 2 2
AB BC AC
AB AB AB
     
 
     
     
i.e., 1 + tan
2
A = sec
2
A
Similarly, on dividing (1) by BC
2
, we get cot
2
A + 1 = cosec
2
A.
By using above identities, we can express each trigonometric ratio in terms of another
ratio. If we know the value of a ratio, we can find all other ratios by using these identities.
THINK AND DISCUSS
Are these identities true for 0
0
< A < 90
0
? If not, for which values of A they are true ?
l sec
2
A  tan
2
A = 1 l cosec
2
A  cot
2
A = 1
DO THIS
(i)If sin C =
15
,
17
then find cos A.(ii) If tan x =
5
,
12
then find sec x.
(iii)If cosec  =
25
7
, then find cot x.
TRY THESE
Evaluate the following and justify your answer.
(i)
2 o 2 o
2 o 2 o
sin 15 sin 75
cos 36 cos 54


(ii)sin 5
o
cos 85
o
+ cos5
o
sin 85
o
(iii)sec 16
o
cosec 74
o
 cot 74
o
tan 16
o
.
Example-13. Show that cot  + tan  = sec  cosec .
Solution : LHS = cot  + tan 
cos sin
sin cos
 
 
 
(why ?)
2 2
cos sin
sin cos
  

 

296. Trigonometry2 8 9
Free Distribution by A.P. Government
1
sin cos

 
(why ?)
1 1
cosec sec
sin cos
   
 
Example-14. Show that tan
2
 + tan
4
 = sec
4
 sec
2

Solution : L.H.S. tan
2
 + tan
4

= tan
2
 (1 + tan
2
)
= tan
2
 . sec
2
 (Why ?)
= (sec
2
  1) sec
2
 (Why ?)
= sec
4
  sec
2
 = R.H.S
Example-15. Prove that
1 cos
1 cos
 
 
= cosec  + cot 
Solution : LHS
1 cos
1 cos
 
 
(multiply and divide by 1 + cos )
1 cos 1 cos
.
1 cos 1 cos
   

   
2
2
(1 cos )
1 cos
 

 
2
2
(1 cos )
sin
 


(Why ?)
1 cos
sin
 


1 cos
cosec cot R.H.S.
sin sin

      
 
EXERCISE 11.4
1. Evaluate the following :
(i)(1 + tan  + sec ) (1 + cot  cosec )
(ii)(sin  + cos )
2
+ (sin   cos )
2
(iii)(sec
2
  1) (cosec
2
 1)

297. Mathematic2 9 0
Free Distribution by A.P. Government
2. Show that (cosec   cot )
2
=
1 cos
1 cos
 
 
3. Show that
1 sin A
secA tan A
1 sin A

 

4. Show that
2
2
2
1 tan A
tan A
cot A 1



5. Show that
1
cos tan .sin
cos
    

6. Simplify secA (1 sinA) (secA + tanA)
7. Prove that (sinA + cosec A)
2
+ (cosA + secA)
2
= 7 + tan
2
A + cot
2
A
8. Simplify (1  cos) (1 +cos) (1 + cot
2
)
9. If sec + tan  = p, then what is the value of sec  tan  ?
OPTIONAL EXERCISE
[This exercise is not meant for examination]
1. Prove that
cot cos cosec 1
cot cos cosec 1
    

    
2. Prove that
sin cos 1 1
sin cos 1 sec tan
   

      
using the identify sec
2
= 1 + tan
2
.
3. Prove that (cocsec A sin A) (sec A cos A) =
1
tan A cot A
4.
2
1 secA sin A
secA 1 cosA



. Prove it.
5.
22
2
2
1 tan A 1 tan A
tan A
1 cotA1 cot A
    
   
  
6.
(sec 1) (1 cos )
(sec 1) (1 cos )
A A
A A
    
   
    
  

298. Trigonometry2 9 1
Free Distribution by A.P. Government
WHAT WE HAVE DISCUSSED
1. In a right angle triangle ABC, right angle is at B,
Side oppositeto angle A
sin A
Hypotenuse
 ,
Side adjacent to angle A
cosA
Hypotenuse

2.
1 1 1 sin A
cosecA ; secA ; tan A , tan A
sin A cosA cot A cosA
   
3. If one of the trignometric ratios of an acute angle is known, the remaining trignometric
ratios of the angle can be determined.
4. The trignometric ratios for angle 0
o
, 30
o
, 45
o
, 60
o
and 90
o
.
5. The value of sin A or cos A never exceeds 1, whereas the value of sec A or cosec A is
always greater than or equal to 1.
6. sin (90
o
 A) = cos A, cos (90
o
 A) = sin A
tan (90
o
 A) = cot A, cot (90
o
 A) = tan A
sec A (90
o
 A) = cosec A, cosec (90
o
 A) = sec A
7. sin
2
A + cos
2
A = 1
sec
2
A  tan
2
A = 1 for 0
o
< A < 90
o
cosec
2
A = 1+ cot
2
A for 0
o
< A < 90
o

300. 2 9 2
Free Distribution by A.P. Government
Mathematic
12.1 INTRO DUCTION
You have studied in social studies that the highest mountain peak in the world is Mount
Everest and its height is 8848 meters.
Kuntala waterfall in Adilabad district is the highest natural waterfall in Andhra Pradesh. Its
height is 147 feet.
How were these heights measured? Can you measure the height of your school building or
the tallest tree in or around your school?
Let us understand through some examples. Vijaya
wants to find the height of a palm trees. She tries to locate the
top most point of the trees. She also imagines a line joining
the top most point and her eye.
This line is called “line of sight”. She also imagines a
horizontal line, parallal to earth, from her eye to the tree.
Here, “the line of sight”, “horizontal line” and “the tree”
form a right angle triangle.
To find the height of the tree, she needs to find a side
and an angle in this triangle.
“The line of sight is above the horizontal line and angle
between the line of sight and the horizontal line is called angle
of elevation”.
Some Applications of
Trigonometry12

L
in
e
o
f s
ig
h
t
Angle of
elevation

301. 2 9 3
Free Distribution by A.P. Government
Some Applications of Trigonometry
Suppose you are standing
on the top of your school
building and you want to find the
distance of borewell from the
building on which you are
standing. For that, you have to
observe the base of the
borewell.
Then, the line of sight from
your eye to the base of borewell
is below the horizontal line from your eye.
Here, “the angle between the line of sight and horizantal line is called angle of depression.”
Trigonometry has been used by surveyors for centuries. They use Theodolites to measure
angles of elevation or depression in the process of survey. In nineteenth century, two large
Theodolites were built by British India for the surveying project “great trigonometric survey”.
During the survey in 1852, the highest mountain peak in the world was discovered in the
Himalayas. From the distance of 160 km, the peak was observed from six different
stations and the height of the peak was calculated. In 1856, this peak was named after Sir
George Everest, who had commissioned and first used the giant Theodolites. Those
theodolites are kept in the museum of the Survey of India in Dehradun for display.
12.2 DRAWING FIGURES TO SOLVE PROBLEMS
When we want to solve the problems of heights and distances, we should consider the
following:
(i)All the objects such as towers, trees, buildings, ships, mountains etc. shall be considered
as linear for mathematical convenience.
(ii)The angle of elevation or angle of depression is considered with reference to the horizontal
line.
(iii)The height of the observer is neglected, if it is not given in the problem.
When we try find heights and distances at an angle of elevation or depression, we need to
visualise geometrically. To find heights and distances, we need to draw figures and with the help
of these figures we can solve the problems. Let us see some examples.
Example-1. The top of a clock tower is observed at angle of elevation of  and the foot of the
tower is at the distance, of d meters from the observer. Draw the diagram for this data.

Horizontal line
Angle of depression
L
i
n
e

o
f

s
i
g
h
t

302. 2 9 4
Free Distribution by A.P. Government
Mathematic


Solution : The diagrams are as shown below :
Example-2. Rinky observes a flower on the ground from the balcony of the first floor of a
building at an angle of depression, . The height of the first floor of the building is x meters. Draw
the diagram for this data.
Solution :
Example-3. A large balloon has been tied with a rope and it is floating in the air. A person has
observed the balloon from the top of a building at angle of elevation of 
1
and foot of the rope at
an angle of depression of 
2
. The height of the building is h feet. Draw the diagram for this data.
Solution : We can see that
BDA = DAE.
d

d

A
B C


B
x
C
AD






2
1
2
E
D
A
B
C
h

303. 2 9 5
Free Distribution by A.P. Government
Some Applications of Trigonometry
DO THIS
1.Draw diagram for the following situations :
(i)A person is flying a kite at angle of elevation  and the length of thread from his
hand to kite is ‘’.
(ii)A person observes two banks of a river at angles of depression 
1
and 
2
(
1
<
2
) from the top of a tree of height h which is at a side of the river. The
width of the river is ‘d’.
THINK AND DISCUSS
1. You are observing top of your school building at an angle of elevation  from a point
which is at d meter distance from foot of the building.
Which trigonometric ratio would you like to consider to find the height of the building?
2. A ladder of length x meter is leaning against a wall making angle  with the ground.
Which trigonometric ratio would you like to consider to find the height of the point on the
wall at which the ladder is touching?
Till now, we have discussed how to draw diagrams as per the situations given. Now, we
shall discuss how to find heights and distances.
Example-4. A boy observed the top of an electric pole to be at an angle of elevation of 60º
when the observation point is 8 meters away from the foot of the pole. Find the height of the pole.
Solution : From the figure, in triangle OAB
OB = 8 meters
AOB = 60º
Let, height of the pole = AB = h meters
(we know the adjacent side and we need to find the opposite side of AOB in the triangle
OAB. Hence we need to consider the trigonometric ratio “tan” to solve the problem).
tan 60º =
AB
OB
3
8
h
 8 3 .h m
O
B
A
60
º
8 m
h

304. 2 9 6
Free Distribution by A.P. Government
Mathematic
Example-5. Rajender observes a person standing on the ground from a helicopter at an angle of
depression 45º. If the helicopter flies at a height of 50 meters from the ground, what is the
distance of the person from Rajender?
Solution : From the figure, in triangle OAB
OA = 50 meters
POB = OAB = 45º (why ?)
OB = distance of the person from Rajender.
(we know the opposite side of OBA and we need to find hypotenuse OB in the triangle OAB.
Hence, we need to consider the ratio “sin”.
sin 45º =
OA
OB
1 50
2x

50 2x meters
EXERCISE - 12.1
1. A tower stands vertically on the ground. From a point which is 15 meter away from the
foot of the tower, the angle of elevation of the top of the tower is 45º. What is the height
of the tower?
2. A tree breaks due to storm and the broken part bends so that the top of the tree touches
the ground by making 30º angle with the ground. The distance between the foot of the
tree and the top of the tree on the ground is 6m. Find the height of the tree before falling
down.
3. A contractor wants to set up a slide for the children to play in the park. He wants to set
it up at the height of 2 m and by making an angle of 30º with the ground. What should be
the length of the slide?
4. Length of the shadow of a 15 meter high pole is 5 3 meters at 7 o’clock in the morning.
Then, what is the angle of elevation of the Sun rays with the gound at the time?
5. You want to erect a pole of height 10 m with the support of three ropes. Each rope has
to make an angle 30º with the pole. What should be the length of the rope?
P O
B
A
45
º
50 m
45
º

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Some Applications of Trigonometry
6. Suppose you are shooting an arrow from the top of a building at an height of 6 m to a
target on the ground at an angle of depression of 60º. What is the distance between you
and the object?
7. An electrician wants to repair an electric connection on a pole of height 9 m. He needs
to reach 1.8 m below the top of the pole to do repair work. What should be the length
of the ladder which he should use, when he climbs it at an angle of 60º with the ground?
What will be the distance between foot of the ladder and foot of the pole?
8. A boat has to cross a river. It crosses the river by making an angle of 60º with the bank
of the river due to the stream of the river and travels a distance of 600m to reach the
another side of the river. What is the width of the river?
9. An observer of height
1.8 m is13.2 m away
from a palm tree. The
angle of elevation of
the top of the tree from
his eyes is 45º. What
is the height of the
palm tree?
10.In the adjacent figure,
AC = 6 cm, AB = 5 cm and BAC = 30º. Find the area of the triangle.
12.3 SOLUTION FOR TWO RIGHT ANGLE TRIANGLES
We have discussed the
solution of a one triangle problem.
What will be the solution if there
are two triangles?
Suppose you are standing
on one side of a tree. You want to
find the height of a tree and you
want to observe the tree from
different line of sights. Then, in this
case, what will be the difference
between the angles?
How can you do this?
Suppose you are observing the
top of the palm tree at an angle of
elevation 45º. The angle of
45º30º
A
C
B
D
E
1
E
B
30º
6 cm
5 cm
C
B

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Mathematic
elevation changes to 30º when you move 11 m away from the tree.
Let us see how we can find height of the tree.
From figure, we have
AB = 11 m
DAC = 30º
DBC = 45º
Let the height of the palm tree CD = h meters
and length of BC = x
AC = 11 + x
from triangle BDC
tan 45º =
DC
BC
1
3
3
h
x h
x
  
...(1)
from triangle ADC
tan 30º =
DC
AC
1
113
h
x


11
3
x
h


11
3 3
h
h 

11
3 3
h
h 
 3 1
11
3 3
h


 
11
3 1
h

meters.
Note :Total height of the palm tree is CD + CE
1
where CE
1
= AE
1
, which is the height of the
girl.
D
B
A
30º
h
45º
C
11 m

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Some Applications of Trigonometry
Example-6. Two men on either side of a temple of 30 meter height observe it at the angles of
elevation 30º and 60º respectively. Find the distance between the two men.
Solution : Height of the temple BD = 30 meter.
Angle of elevation of one person BAD = 30º
Angle of elevation of another person BCD = 60º
Let the distance between the first person and the temple, AD = x and distance between the
second person and the temple, CD = d
From BAD From BCD
tan 30º =
BD
AB
tan 60º =
BD
d
1 30
3x

30
3
d

30 3x .......... (1)
30
3
d
.......... (2)
from (1) and (2) distance between the persons = BC + BA = x + d
30 30 4 120
30 3 40 3
3 3 3

    
meter
Example-7. A straight highway leads to the foot of a tower. A man standing at the top of the
tower observes a car at an angle of depression 30º. The car is approaching the foot of the tower
with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60º.
Find the time taken by the car to reach the foot of the tower from this point.
Solution :
Let the distance travelled by the car in 6 seconds = AB = x meters
Heights of the tower CD = x meters
The remaining distance to be travelled by the carBC = d meters
and AC = AB + BC = (x + d) meters
PDA = DAP = 30º (why?)
PDB = DBP = 60º (why?)
From BCD
x
30º 60º
d
A
B
C
D

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Mathematic
P
D
B
A
30º
h
60º
C
d
30º
60º
CD
tan 60º
BC

3
h
d

3h d ...(1)
From ACD
CD
tan30º
AC

1
( )3
h
x d


( )
3
x d
h


...(2)
From (1) & (2), we have
3
3
x d
d


x + d = 3d
x = 2d
2
x
d
Time taken to travel ‘x’ meters = 6 seconds.
Time taken to travel the distance of ‘d’ meters
i.e.,
2
x
meters = 3 seconds.
EXERCISE - 12.2
1. A TV tower stands vertically on the side of a road. From a point on the other side
directly opposite to the tower, the angle of elevation of the top of tower is 60
0
. From
another point 10 m away from this point, on the line joining this point to the foot of the
tower, the angle of elevation of the top of the tower is 30º. Find the height of the tower
and the width of the road.

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Some Applications of Trigonometry
2. A 1.5 m tall boy is looking at the top of a temple which is 30 meter in height. The angle
of elevation from his eye to the top of the crown of the temple increases from 30º to 60º
as he walks towards the temple. Find the distance he walked towards the temple.
3. A statue stands on the top of a 2m tall pedestal. From a point on the ground, the angle
of elevation of the top of the statue is 60º

and from the same point, the angle of elevation
of the top of the pedestal is 45º. Find the height of the statue.
4. From the top of a building, the angle of elevation of the top of a cell tower is 60º and the
angle of depression to its foot is 45º. If distance of the building from the tower is 7m,
then find the height of the tower.
5. A wire of length 18 m had been tied with electric pole at an angle of elevation 30º with
the ground. Because it was convering a long distance, it was cut and tied at an angle of
elevation 60º with the ground. How much length of the wire was cut?
6. The angle of elevation of the top of a building from the foot of the tower is 30º and the
angle of elevation of the top of the tower from the foot of the building is 60º. If the tower
is 30 m high, find the height of the building.
7. Two poles of equal heights are standing opposite to each other on either side of the road,
which is 120 feet wide. From a point between them on the road, the angles of elevation
of the top of the poles are 60º and 30º respectively. Find the height of the poles and the
distances of the point from the poles.
8. The angles of elevation of the top of a tower from two points at a distance of 4 m and
9 m, from the base of the tower and in the same straight line with it are complementary.
Prove the height of the tower is 6 m.
9. The angle of elevation of a jet plane from a point A on the ground is 60º. After a flight of
15 seconds, the angle of elevation changes to 30º. If the jet plane is flying at a constant
height of 15003 meter, find the speed of the jet plane.  3 1.732
10.Ganesh observes the top and foot of an opposite building from a window at first floor of
his house which is at 8 meter height. The angles of elevation and depression of the top
and the foot of the building are 60º and 30º respectively. Find the height of the building?

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Mathematic
OPTIONAL EXERCISE
[This exercise his not meant for examination]
1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the
top of a vertical pole to the ground. Find the height of the pole, if the angle made by the
rope with the ground level is 30º.
2. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of
depression of two ships are 30º and 45º. If one ship is exactly behind the other on the
same side of the lighthouse, find the distance between the two ships.
3. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of
88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at
any instant is 60º. After some time, the angle of elevation reduces to 30º. Find the distance
travelled by the balloon during the interval.
4. A straight highway leads to the foot of a tower. A man standing at the top of the tower
observes a car at an angle of depression of 30º, which is approaching the foot of the
tower with a uniform speed. Six seconds later, the angle of depression of the car is found
to be 60º. Find the time taken by the car to reach the foot of the tower from this point.
5. The angles of elevation of the top of a tower from two points at a distance of 4 m and
9 m from the base of the tower and in the same straight line with it are complementary.
Prove that the height of the tower is 6 m.
WHAT WE HAVE DISCUSSED
In this chapter, we have studied the following points :
1.(i)The line of sight is the line drawn from the eye of an observer to a point on the object
being viewed by the observer.
(ii)The angle of elevation of the object viewed, is the angle formed by the line of sight
with the horizontal when it is above the horizontal level, i.e., the case when we raise
our head to look at the object.
(iii)The angle of depression of an object viewed, is the angle formed by the line of sight
with the horizontal when it is below the horizontal level, i.e., the case when we lower
our head to look at the object.
2.The height or length of an object or the distance between two distant objects can be
determined with the help of trigonometric ratios.

312. 13.1 INTRO DUCTION
Kumar and Sradha were walking together to play a carroms match:
Kumar:Do you think we will win?
Sradha:There are 50 percent chances of that. We may win.
Kumar:How do you say 50 percent?
Do you think Sudha is right in her statement?
Is her chance of wining 50%?
In this chapter, we will think about such questions. We will also discuss words like
‘probably’, ‘likely’, ‘possibly’, etc. and how to quantify these. In class VIII we talked about
events that are extremely likely and in fact, are almost certain and those that are extremely unlikely
and hence almost impossible. We also talked about chance, luck and the fact that an event occurs
one particular time does not mean that it would happen each time. In this chapter, we would try
to learn how the likelihood of an event can be quantified.
This quantificatioin into a numerical measure is referred to as finding 'Probability'.
13.1.1 WHAT IS PROBABILITY
Consider an experiment: A normal coin was tossed 1000 times. Head turned up 455 times
and tail turned up 545 times. If we try to find the likelihood of getting heads we may say it is 455
out of 1000 or
455
1000
or 0.455.
This estimation of probability is based on the results of an actual
experiment of tossing a coin 1000 times. These estimates are called experimental
or empirical probabilities. In fact, all experimental probabilities are based on
the results of actual experiments and an adequate recording of what happens
in each of the events. These probabilities are only 'estimates'. If we perform
the same experiment for another 1000 times, we may get slightly different
data, giving different probability estimates.
Probability13

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Many other persons from different parts of the world have done this kind of experiment
and recorded the number of heads that turned up.
For example, the eighteenth century French naturalist Comte de Buffon, tossed a coin
4040 times and got 2048 heads. The experimental probability of getting a head, in this case, was
2048
4040
i.e., 0.507.
J.E. Kerrich, from Britain, recorded 5067 heads in 10000 tosses of a coin. The experimental
probability of getting a head, in this case, was
5067
10000
= 0.5067. Statistician Karl Pearson spent
some more time, making 24000 tosses of a coin. He got 12012 heads, and thus, the experimental
probability of a head obtained by him was 0.5005.
Now, suppose we ask, 'What will be the experimental probability of getting a head, if the
experiment is carried on up to, say, one million times? Or 10 million times? You would intuitively
feel that as the number of tosses increases, the experimental probability of a head (or a tail) may
settle down closer and closer to the number 0.5 , i.e.,
1
2
. This matches the theoretical probability
of getting a head (or getting a tail), we will learn how to find the theoretical probability.
This chapter is an introduction to the theoretical (also called classical) probability of an
event, we will discuss simple problems based on this concept.
13.2 PROBABILITY - A THEORETICAL APPROACH
Let us consider the following situation: Suppose a ‘fair’ coin is tossed at random.
When we speak of a coin, we assume it to be 'fair', that is, it is symmetrical so that there is
no reason for it to come down more often on one side than the other. We call this property of the
coin as being 'unbiased'. By the phrase 'random toss', we mean that the coin is allowed to fall
freely without any bias or interference. (Here we dismiss the possibility of its 'landing' on its edge,
which may be possible, for example, if it falls on sand). We refer to this by saying that the
outcomes, head and tail, are equally likely.
For basic understanding of probability, in this chapter, we will assume that all the experiments
have equally likely outcomes.
Now, we know that the experimental or empirical probability P(E) of an event E is
P(E) =
Number of trials in which the event happened
Total number of trials

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DO THESE
a.Outcomes of which of the following experiments are equally likely?
1.Getting a digit 1, 2, 3, 4, 5 or 6 while throwing a dice.
2.Picking a different colour ball from a bag of 5 red balls, 4 blue balls and 1 black ball.
3.Winning in a game of carrom.
4.Units place of a two digit number selected may be 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9.
5.Picking a different colour ball from a bag of 10 red balls, 10 blue balls and 10 black
balls.
6.Raining on a particular day of July.
b. Are the outcomes of every experiment equally likely?
c.Give examples of 5 experiments that have equally likely outcomes and five more that do
not have equally likely outcomes.
ACTIVITY
(i)Take any coin, toss it ten times, 50 times, 100 times, 150 times and count the number of
times a head and a tail come up seperately. Record your observations in the following
table:-
S.Number ofNumber ofProbability ofNumber ofProbability of
No.experiments heads head tails tails
1. 50
2. 100
3. 150
What do you observe? Obviously, as the number of experiments are more and more,
probability of head or tail reaches 50% or
1
2
. This empirical interpretation of probability can be
applied to every event associated with an experiment that can be repeated a large number of
times.
Probability and Modelling
The requirement of repeating an experiment has some limitations, as it may be very
expensive or unfeasible in many situations. Of course, it worked well in coin tossing or die
throwing experiments. But how about repeating the experiment of launching a satellite in

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order to compute the empirical probability of its failure during launching, or the repetition of
the phenomenon of an earthquake to compute the empirical probability of a multi-storeyed
building getting destroyed in an earthquake? For finding these probabilities we calculate models
of behaviour and use them to estimate behaviour and likely outcomes. Such models are complex
and are validated by predictions and outcomes. Forecast of weather, result of an electionj,
population demography, earthquakes, crop production etc. are all based on such models and
their predictions.
The assumption of equally likely outcomes (which is valid in many experiments, as in two
of the examples seen, of a coin and of a die) is one of the assumption that leads us to the following
definition of probability of an event.
The theoretical probability (also called classical probability) of an event T, written as P(T),
is defined as
P(T) =
Number of outcomes favourable to E
Number of all possible outcomes of the experiment
where we assume that the outcomes of the experiment are equally likely. We usually simply
refer to theoretical probability as Probability.
13.3 MUTUALLY EXCLUSIVE EVENTS
If a coin is tossed, we get head or tail, but not both. Similarly, if we select a student of a
high school that student may belong to one of either 6, 7, 8, 9 or 10 class, but not to any two or
more classes. In both these examples, occurrence of an event prevents the occurrence of other
events. Such events are called mutually exclusive events.
Two or more events of an experiment, where occurance of an event is prevents all other
events are called Mutually Exclusive Events. We will discuss this in more detail later in the
chapter.
This definition of probability was given by Pierre Simon Laplace
in 1795.
Probability theory had its origin in the16
th
century when an Italian
physician and mathematician J. Cardan wrote the first book on the subject,
The Book on Games of Chance. James Bernoulli (1654 -1705), A. de
Moivre (1667-1754), and Pierre Simon Laplace are among those who
made significant contributions to this field. In recent years, probability
has been used extensively in many areas such as biology, economics,
genetics, physics, sociology etc.
Pierre Simon Laplace
(1749 – 1827)

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13.4.1 FINDING PROBABILITY
How do we find the probability of events that are equally likely. We consider the tossing of
a coin as an event associated with experiments where the equally likely assumption holds. In
order to proceed, we recall that there are two possible outcomes each time. This is the called the
sample space. We can say that the sample space of one toss is {H, T}. For the experiement of
drawing out a ball from a bag containing red, blue, yellow and white ball, the sample space is {R,
B, Y, W}. What is the sample space for the throw of a dice?
DO THIS
Think of more situations with equally likely events and find the sample space.
Let us now try to find the probability of equally likely events that are mutually exclusive.
Example-1.Find the probability of getting a head when a coin is tossed once. Also find the
probability of getting a tail.
Solution : In the experiment of tossing a coin once, the number of possible outcomes is two -
Head (H) and Tail (T). Let E be the event 'getting a head'. The number of outcomes favourable
to E, (i.e., of getting a head) is 1. Therefore,
P(E) = P (head) =
Number of outcomes favourable to E
Number of all possible outcomes
=
1
2
Similarly, if F is the event 'getting a tail', then
P(F) = P(tail) =
1
2
(Guess why?)
Example-2.A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the
same size. Kritika takes out a ball from the bag without looking into it. What is the probability
that she takes a (i) yellow ball? (ii) red ball? (iii) blue ball?
Solution : Kritika takes out a ball from the bag without looking into it. So, it is equally likely that
she takes out any one of them.
Let Y be the event 'the ball taken out is yellow', B be the event 'the ball taken out is blue',
and R be the event 'the ball taken out is red'.
Now, the number of possible outcomes = 3.
(i)The number of outcomes favourable to the event Y = 1.
So, P(Y) =
1
3
. Similarly, P(R) =
1
3
and P(B) =
1
3

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Remarks
1.An event having only one outcome in an experiment is called an elementary event. In
Example 1, both the events E and F are elementary events. Similarly, in Example 2, all
the three events, Y, B and R are elementary events.
2.In Example 1, we note that : P(E) + P(F) = 1
In Example 2, we note that : P(Y) + P(R) + P(B) = 1
3.In events like a throw of dice, probability of getting less than 3 and of getting a 3 or
more than three are not elementary events of the possible outcomes. In tossing two
coins {HH}, {HT}, {TH} and {TT} are elementary events. If we find the probability
of all the elementary events and add them, we would get the total as 1.
Can you see that the sum of the probabilities of all the elementary events of an
experiment is 1?
Example-3.Suppose we throw a die once. (i) What is the probability of getting a number
greater than 4? (ii) What is the probability of getting a number less than or equal to 4?
Solution : (i) In rolling an unbaised dice
Sample space S= {1, 2, 3, 4, 5, 6}
No. of events n(S)= 6
Favourable event for numberE= {5, 6}
greater than 4
No. of favourable eventsn(E)= 2
Probability P(E)=
2
6
=
1
3
Note : Observe taht in example 3, the event’s are not elementary events.
(ii) Let F be the event 'getting a number less than or equal to 4'.
Sample space S= {1, 2, 3, 4, 5, 6}
No. of events n(S)= 6
Favourable event for number lessF= {1, 2, 3, 4}
or equal to 4
No. of favourable eventsn(F)= 4
Probability P(F)=
4
6
=
2
3

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Are the events E and F in the example above elementary events?
No, they are not elementary events. The event E has 2 outcomes and the event F has 4
outcomes.
13.4.2 EVENTS AND PROBAB ILITY
In the previous section we read about elementary events. Then in example-3, we calculated
probability of events which are not elementary. We saw,
P(E) + P(F) =
1
3
+
2
3
= 1
Here F is the same as 'not E' because there are only two events.
We denote the event 'not E' by E. This is called the complement event of event E.
So, P(E) + P(not E) = 1
i.e., P(E) + P(E) = 1, which gives us P(E) = 1 - P(E).
In general, it is true that for an event E, P() = 1 – P(E)
DO THIS
(i)Is getting a head complementary to getting a tail? Give reasons.
(ii)Is getting a 1 complementary to getting 2, 10, 4, 5, 6? Give reasons for your answer.
(iii)Think of five new pair of events that are complementary.
13.4.3 IM POSSIBLE AND CERTAIN EVENTS
Consider the following about the throws of a die with sides marked as 1, 2, 3, 4, 5, 6.
(i)What is the probability of getting a number 7 in a single throw of a die?
We know that there are only six possible outcomes in a single throw of this die. These
outcomes are 1, 2, 3, 4, 5 and 6. Since no face of the die is marked 7, there is no outcome
favourable to 7, i.e., the number of such outcomes is zero. In other words, getting 7 in a single
throw of a die, is impossible.
So P(getting 7)=
0
6
= 0
That is, the probability of an event which is impossible to occur is 0. Such an event is
called an impossible event.
(ii) What is the probability of getting 6 or a number less than 6 in a single throw of a die?

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Since every face of a die is marked with 6 or a number less than 6, it is sure that we will
always get one of these when the dice is thrown once. So, the number of favourable outcomes is
the same as the number of all possible outcomes, which is 6.
Therefore, P(E) = P(getting 6 or a number less than 6) =
6
6
= 1
So, the probability of an event which is sure (or certain) to occur is 1. Such an event is
called a sure event or a certain event.
Note : From the definition of probability P(E), we see that the numerator (number of outcomes
favourable to the event E) is always less than or equal to the denominator (the number of all
possible outcomes). Therefore, 0  P(E)  1.
TRY THESE
1.A childhas a die whose six faces show the letters A, B, C, D, E and F. The die is thrown
once. What is the probability of getting (i) A? (ii) D?
2.Which of the following cannot be the probabiliyt of an event?
(a) 2.3 (b)-1.5 (c)15% (D)0.7
THINK AND DISCUSS
1.Why is tossing a coin considered to be a fair way of deciding which team should get the ball
at the beginning of any game?
2.Can
7
2
be the probability of an event? Explain.
3.Which of the following arguments are correct and which are not correct? Give reasons.
i)If two coins are tossed simultaneously there are three possible outcomes - two heads, two
tails or one of each. Therefore, for each of these outcomes, the probability is
1
3
.
ii)If a dice is thrown, there are two possible outcomes - an odd number or an even number.
Therefore, the probability of getting an odd number is
1
2
.
13.5 DECK OF CARDS AND PROBABILITY
Have you seen a deck of playing cards?
A deck of playing cards consists of 52 cards which are divided into 4 suits of 13 cards
each. They are black spades (), red hearts (), red diamonds () and black clubs ().

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The cards in each suit are Ace, King, Queen, Jack, 10, 9, 8, 7, 6, 5, 4, 3 and 2. Kings,
Queens and Jacks are called face cards. Many games are played with this deck of cards, some
games are played with part of
the deck and some with two
decks even. The study of
probability has a lot to do with
card and dice games as it helps
players to estimate possibilities
and predict how the cards could
be distributed among players.
Example-4. One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability
that the card will be an ace, (ii) not be an ace.
Solution : Well-shuffling ensures equally likely outcomes.
(i)There are 4 aces in a deck.
Let E be the event 'the card is an ace'.
The number of outcomes favourable to E = 4
The number of possible outcomes = 52 (Why ?)
Therefore, P(E) =
4 1
=
52 13
(ii)Let F be the event 'card drawn is not an ace'.
The number of outcomes favourable to the event F = 52 - 4 = 48 (Why?)
The number of possible outcomes = 52
Therefore, P(F) =
48 12
=
52 13
Alternate Method : Note that F is nothing but E.
Therefore, we can also calculate P(F) as follows:
P (F) = P(E) = 1 - P(E) = 1 -
1 12
=
13 13
TRY THIS
You have a single deck of well shuffled cards. Then,
1.What is the probability the card drawn will be a queen?

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Free Distribution by A.P. Government
2.What is the probability that it is a face card?
3.What is the probability it is a spade?
4.What is the probability that is the face card of spades?
5.What is the probability it is not a face card?
13.6 USE OF PROBABILITY
Let us look at some more occasions where probability may be useful. We know that in
sports some countries are strong and others are not so strong. We also know that when two
players are playing it is not that they win equal times. The probability of winning of the player or
team that wins more often is more than the probability of the other player or team. We also
discuss and keep track of birthdays. It sometimes happens that people we know have the same
birthdays. Can we find out whether this is a common event or would it only happen occassionally.
Classical probability helps us do this.
Example-5.Sangeeta and Reshma, play a tennis match. It is known that the probability of
Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?
Solution : Let S and R denote the events that Sangeeta wins the match and Reshma wins the
match, respectively.
The probability of Sangeeta's winning= P(S) = 0.62 (given)
The probability of Reshma's winning= P(R) = 1 - P(S)
= 1 -0.62 = 0.38 [R and S are complementary]
Example-6.Sarada and Hamida are friends. What is the probability that both will have (i)
different birthdays? (ii) the same birthday? (ignoring a leap year).
Solution : Out of the two friends, one girl, say, Sarada's birthday can be any day of the year.
Now, Hamida's birthday can also be any day of 365 days in the year. We assume that these 365
outcomes are equally likely.
(i) If Hamida's birthday is different from Sarada's, the number of favourable outcomes for her
birthday is 365 - 1 = 364
So, P (Hamida's birthday is different from Sarada's birthday) =
364
365
(ii) P(Sarada and Hamida have the same birthday) = 1 - P (both have different birthdays)
= 1-
364
365
[ Using P(
E
) = 1 - P(E)] =
1
365

322. Probability3 1 3
Free Distribution by A.P. Government
Example-7.There are 40 students in Class X of a school of whom 25 are girls and 15 are
boys. The class teacher has to select one student as a class representative. She writes the name
of each student on a separate card, the cards being identical. Then she puts cards in a bag and
stirs them thoroughly. She then draws one card from the bag. What is the probability that the
name written on the card is the name of (i) a girl? (ii) a boy?
Solution : There are 40 students, and only one name card has to be chosen.
(i) The number of all possible outcomes is 40
The number of outcomes favourable for a card with the name of a girl = 25 (Why?)
 P (card with name of a girl) = P(Girl) =
25 5
=
40 8
(ii) The number of outcomes favourable for a card with the name of a boy = 15 (Why?)
Therefore, P(card with name of a boy) = P(Boy) =
15 3
=
40 8
or P(Boy) = 1 - P(not Boy) = 1 - P(Girl) = 1 -
5 3
=
8 8
EXERCISE - 13.1
1.Complete the following statements:
(i) Probability of an event E + Probability of the event 'not E' = ______________
(ii) The probability of an event that cannot happen is__________.
Such an event is called __________
(iii) The probability of an event that is certain to happen is __________.
Such an event is called______
(iv) The sum of the probabilities of all the elementary events of an experiment is_________
(v) The probability of an event is greater than or equal to __________ and less than or
equal to _______
2.Which of the following experiments have equally likely outcomes? Explain.
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.
(iii) A trial is made to answer a true-false question. The answer is right or wrong.
(iv) A baby is born. It is a boy or a girl.

323. Mathematic3 1 4
Free Distribution by A.P. Government
3.If P(E) = 0.05, what is the probability of 'not E'?
4.A bag contains lemon flavoured candies only. Malini takes out one candy without looking
into the bag. What is the probability that she takes out
(i) an orange flavoured candy? (ii) a lemon flavoured candy?
5.Rahim takes out all the heads from the cards:
i.What is the probability of picking out an ace from the remaining pack.
ii.Picking out a diamonds.
iii.Picking out a card that is not a heart.
iv.Picking out the Ace of hearts.
6.It is given that in a group of 3 students, the probability of 2 students not having the same
birthday is 0.992. What is the probability that the 2 students have the same birthday?
7.A die is thrown once. Find the probability of getting
(i) a prime number; (ii) a number lying between 2 and 6; (iii) an odd number.
8.What is the probability of drawing out a red king from a deck of cards?
9.Make 5 more problem of this kind using dice, cards or birthdays and discuss with friends.
13.7 MORE APPLICATIONS OF PROBABILITY
We have seen some example of use of probability. Think about the contents and ways
probability has been used in these. We have seen again that probability of complementary events
add to 1. Can you identify in the examples and exercises given above, and those that follow,
complementary events and elementary events? Discuss with teachers and friends. Let us see
more uses.
Example-8.A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random
from the box, what is the probability that it will be
(i) white? (ii) blue? (iii) red?
Solution : Saying that a marble is drawn at random means all the marbles are equally likely to be
drawn.
 The number of possible outcomes = 3 +2 + 4 = 9 (Why?)
Let W denote the event 'the marble is white', B denote the event 'the marble is blue' and R
denote the event 'marble is red'.
(i) The number of outcomes favourable to the event W = 2
So, P(W) =
2
9

324. Probability3 1 5
Free Distribution by A.P. Government
Similarly, (ii) P(B) =
3 1
=
9 3
and (iii) P(R) =
4
9
Note that P(W) + P(B) + P(R) = 1.
Example-9.Harpreet tosses two different coins simultaneously (say, one is of D1 and other of
D2). What is the probability that she gets at least one head?
Solution : We write H for 'head' and T for 'tail'. When two coins are tossed simultaneously, the
possible outcomes are (H, H), (H, T), (T, H), (T, T), which are all equally likely. Here (H, H)
means heads on the first coin (say on D1) and also heads on the second coin (D2). Similarly (H,
T) means heads up on the first coin and tail up on the second coin and so on.
The outcomes favourable to the event E, 'at least one head' are (H, H), (H, T) and (T, H).
So, the number of outcomes favourable to E is 3.
P(E) =
3
4
[Since the total possible outcomes = 4]
i.e., the probability that Harpreet gets at least one head is
3
4
Note : You can also find P(E) as follows:
P (E) = 1 - P( E ) = 1 -
1 3
=
4 4
Therefore, P(E) = P(no head) =
1
4
Check This
Did you observe that in all the examples discussed so far, the number of possible outcomes
in each experiment was finite? If not, check it now.
There are many experiments in which the outcome is number between two given numbers,
or in which the outcome is every point within a circle or rectangle, etc. Can you count the number
of all possible outcomes in such cases? As you know, this is not possible since there are infinitely
many numbers between two given numbers, or there are infinitely many points within a circle. So,
the definition of theoretical probability which you have learnt so far cannot be applied in the
present form.
What is the way out? To answer this, let us consider the following example:
Example-10.(Not for examination) In a musical chair game, the person playing the music has
been advised to stop playing the music at any time within 2 minutes after she starts playing. What
is the probability that the music will stop within the first half-minute after starting?
Solution : Here the possible outcomes are all the numbers between 0 and 2. This is the portion
of the number line from 0 to 2

325. Mathematic3 1 6
Free Distribution by A.P. Government
Let E be the event that 'the music is stopped within the first half-minute'.
The outcomes favourable to E are points on the number line from 0 to
1
2
The distance from 0 to 2 is 2, while the distance from 0 to
1
2
is
1
2
Since all the outcomes are equally likely, we can argue that, of the total distance of 2, the
distance favourable to the event E is
1
2
So, P(E) =
Distance favourable to the event E
Total distance in which outcomes can lie
We now try to extend the idea of Example 10 for finding the probability as the ratio of the
favourable area to the total area.
Example-11. A missing helicopter is
reported to have crashed somewhere in the
rectangular region as shown in the figure.
What is the probability that it crashed inside
the lake shown in the figure?
Solution : The helicopter is equally likely
to crash anywhere in the region. Area of
the entire region where the helicopter can
crash = (4.5 × 9) km
2
= 40.5 km
2
Area of the lake = (2.5 × 3) km
2
= 7.5 km
2
Therefore, P (helicopter crashed in the lake) =
7.5 5
=
40.5 27
Example-12.A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4
have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha,
another trader, will only reject the shirts which have major defects. One shirt is drawn at random
from the carton. What is the probability that
(i) it is acceptable to Jimmy?(ii) it is acceptable to Sujatha?
Solution : One shirt is drawn at random from the carton of 100 shirts. Therefore, there are 100
equally likely outcomes.
0 1
2
1 2
6 .km
Lake
9 .km
4
.5

.
k
m
2

.
k
m

326. Probability3 1 7
Free Distribution by A.P. Government
(i) The number of outcomes favourable (i.e., acceptable) to Jimmy = 88 (Why?)
Therefore, P (shirt is acceptable to Jimmy) =
88
100
= 0.88
(ii) The number of outcomes favourable to Sujatha = 88 + 8 = 96 (Why?)
So, P (shirt is acceptable to Sujatha) =
96
100
= 0.96
Example-13. Two dice, one red and one white, are thrown at the same time. Write down all the
possible outcomes. What is the probability that the sum of the two numbers appearing on the top
of the dice is (i) 8? (ii) 13? (iii) less than or equal to 12?
Solution : When the red dice shows '1', the
white dice could show any one of the
numbers 1, 2, 3, 4, 5, 6. The same is true
when the red dice shows '2', '3', '4', '5' or
'6'. The possible outcomes of the experiment
are shown in the figure; the first number in
each ordered pair is the number appearing
on the red dice and the second number is
that on the white dice.
Note that the pair (1, 4) is different from (4, 1). (Why?)
So, the number of possible outcomes = 6 × 6 = 36.
(i) The outcomes favourable to the event 'the sum of the two numbers is 8' denoted by E, are: (2,
6), (3, 5), (4, 4), (5, 3), (6, 2) (See figure)
i.e., the number of outcomes favourable to E = 5.
Hence, P(E) =
5
36
(ii) As there is no outcome favourable to the event F, 'the sum of two numbers is 13',
So, P(F) =
0
36
= 0
(iii) As all the outcomes are favourable to the event G, 'sum of two numbers is 12',
So, P(G) =
36
36
= 1
1, 1
1 2 3 4 5 6
1
2
3
4
5
6
2, 1
3, 1
4, 1
5, 1
6, 1
1, 2
2, 2
3, 2
4, 2
5, 2
6, 2
1, 3
2, 3
3, 3
4, 3
5, 3
6, 3
1, 4
2, 4
3, 4
4, 4
5, 4
6, 4
1, 5
2, 5
3, 5
4, 5
5, 5
6, 5
1, 6
2, 6
3, 6
4, 6
5, 6
6, 6

327. Mathematic3 1 8
Free Distribution by A.P. Government
EXERCISE - 13.2
1.A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What
is the probability that the ball drawn is (i) red ? (ii) not red?
2.A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken
out of the box at random. What is the probability that the marble taken out will be (i) red?
(ii) white ? (iii) not green?
3.A piggy bank contains hundred 50p coins, fifty D1 coins, twenty D2 coins and ten D5 coins.
If it is equally likely that one of the coins will fall out when the bank is turned upside down,
what is the probability that the coin (i) will be a 50 p coin? (ii) will not be a D5 coin?
4.Gopi buys a fish from a shop for his aquarium. The shopkeeper takes
out one fish at random from a tank containing 5 male fish and 8 female
fish (See figure). What is the probability that the fish taken out is a
male fish?
5.A game of chance consists of spinning an arrow which comes t o rest
pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (See figure), and
these are equally likely outcomes. What is the probability that it will
point at
(i) 8 ? (ii) an odd number?
(iii) a number greater than 2?(iv) a number less than 9?
6.One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour (ii) a face card(iii) a red face card
(iv) the jack of hearts (v) a spade(vi) the queen of diamonds
7.Five cards-the ten, jack, queen, king and ace of diamonds, are well-shuffled with their
face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked
up is (a) an ace? (b) a queen?
8.12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look
at a pen and tell whether or not it is defective. One pen is taken out at random from this lot.
Determine the probability that the pen taken out is a good one.
9.A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot.
What is the probability that this bulb is defective? Suppose the bulb drawn in previous
case is not defective and is not replaced. Now one bulb is drawn at random from the rest.
What is the probability that this bulb is not defective?
18
7
6
54
3
2

328. Probability3 1 9
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10.A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random
from the box, find the probability that it bears (i) a two-digit number (ii ) a perfect square
number (iii) a number divisible by 5.
11.Suppose you drop a die at random on the rectangular region
shown in figure. What is the probability that it will land
inside the circle with diameter 1m?
12.A lot consists of 144 ball pens of which 20 are defective
and the others are good. Nuri will buy a pen if it is good,
but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to
her. What is the probability that (i) She will buy it? (ii) She will not buy it ?
13.Two dice are rolled simultaneously and counts are added (i) complete the table given
below:
Event : 'Sum on 2 dice'23456789101112
Probability
1
36
5
36
12
36
(ii) A student argues that 'there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and
12.Therefore, each of them has a probability
1
11
. Do you agree with this argument? Justify
your answer.
14.A game consists of tossing a one rupee coin 3 times and noting its outcome each time.
Hanif wins if all the tosses give the same result i.e., three heads or three tails, and loses
otherwise. Calculate the probability that Hanif will lose the game.
15.A dice is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5
will come up at least once? [Hint : Throwing a die twice and throwing two dice
simultaneously are treated as the same experiment].
OPTIONAL EXERCISE
[This exercise is not meant for examination]
1.Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday
to Saturday). Each is equally likely to visit the shop on any day as on another day. What is
the probability that both will visit the shop on (i) the same day? (ii) consecutive days? (iii)
different days?
2.A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is
double that of a red ball, determine the number of blue balls in the bag.
3 .m
2

.
m

329. Mathematic3 2 0
Free Distribution by A.P. Government
3.A box contains 12 balls out of which x are black. If one ball is drawn at random from the
box, what is the probability that it will be a black ball? If 6 more black balls are put in the
box, the probability of drawing a black ball is now double of what it was before. Find x.
4.A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at
random from the jar, the probability that it is green is 23. Find the number of blue balls in
the jar.
WHAT WE HAVE DISCUSSED
In this chapter, you have studied the following points:
1.We have looked at experimental probability and theoretical probability.
2.The theoretical (classical) probability of an event E, written as P(E), is defined as
P (E) =
Number of trails in which the event happened
Total number of trails
where we assume that the
outcomes of the experiment are equally likely.
3.The probability of a sure event (or certain event) is 1.
4.The probability of an impossible event is 0.
5.The probability of an event E is a number P(E) such that 0  P (E)  1
6.An event having only one outcome is called an elementary event. The sum of the probabilities
of all the elementary events of an experiment is 1.
7.For any event E, P (E) + P (
E
) = 1, where
E
stands for 'not E'. E and
E
are called
complementary events.
8.Some more terms used in the chapter are given below:
Equally likely events:Two or more events are said to be equally likely if each one of
them has an equal chance of occurrance.
Mutually Exclusive events: Two or more events are mutually exclusive if the occurrence of
each event prevents the every other event.
Complementary events :Consider an event has few outcomes. Event of all other outcomes
in the sample survey which are not in the favourable event is
called Complementary event.
Exhaustive events :All the events are exhaustive events if their union is the sample
space.
Sure events :The sample space of a random experiment is called sure or certain
event as any one of its elements will surely occur in any trail of
the experiment.
Impossible event :An event which will occur on any account is called an impossible
event.

330. 14.1 INTRO DUCTIO N
Ganesh had recorded the marks of 26 children in his class in the mathematics half yearly
exam in the register as follows:
Table 14.1
Arjun 76 Narayan 12
Kamini 82 Suresh 24
Shafik 64 Durga 39
Keshar 53 Shiva 41
Lata 90 Raheem 69
Rajendra 27 Rama 73
Ram 34 Kartik 94
Sudha 74 Joseph 89
Krishna 76 Ikram 64
Som 65 Laxmi 46
Gauri 47 Seeta 19
Upendra 54 Rehana 53
Ramaiah 36 Anne 69
Is the data given in the list above organized? Why or why not?
His teacher asked him to report on how his class students had performed in mathematics
in their half yearly exam.
Statistics14

331. Mathematic3 2 2
Free Distribution by A.P. Government
Ganesh made the following table to understand the performance of his class:
Marks Number of children
0 - 33 4
34 - 50 6
51 - 75 10
76 - 100 6
Is the data given in the above table grouped or ungrouped?
He showed this table to his teacher and the teacher quickly understood how the class
had performed. We can see that most children have got marks between 51-75. Do you think
Ganesh should have used smaller range? Why or why not?
In the previous class, you had learnt about the difference between grouped and ungrouped
data as well as how to present this data in the form of tables. You had also learnt to calculate the
mean value for ungrouped data. Let us recall this learning and then learn to calculate the mean,
median and mode for grouped data.
14.2 MEAN OF UNGROUPED DATA
As we know the mean (or average) of observations is the sum of the values of all the
observations divided by the total number of observations. Let x
1
, x
2
,. . ., x
n
be observations with
respective frequencies f
1
, f
2
, . . ., f
n
. This means that observation x
1
occurs f
1
times, x
2
occurs f
2
times, and so on.
Now, the sum of the values of all the observations = f
1
x
1
+ f
2
x
2
+ . . . + f
n
x
n
, and the
number of observations = f
1
+ f
2
+ . . . + f
n
.
So, the mean x of the data is given by
1 1 2 2
1 2
... ... ...
... ... ...
n n
n
f x f x f x
x
f f f
  

  
Recall that we can write this in short using the Greek letter  which means summation
i.e.,
i i
i
f x
x
f



Example-1. The marks obtained in mathematics by 30 students of Class X of a certain school
are given in table below. Find the mean of the marks obtained by the students.
Marks obtained (x
i
) 10 20 36 40 50 56 60 70 72 80 88 92 95
Number of student ( f
i
) 1 1 3 4 3 2 4 4 1 1 2 3 1

332. Statistics3 2 3
Free Distribution by A.P. Government
Let us organize this data in the form of a table 14.2
Table 14.2
Marks Number of
i i
f x
obtained (x
i
)students ( fi)
10 1 10
20 1 20
36 3 108
40 4 160
50 3 150
56 2 112
60 4 240
70 4 280
72 1 72
80 1 80
88 2 176
92 3 276
95 1 95
Total f=30
i i
f x = 1779
So,
1779
59.3
30
i i
i
f x
x
f

  

Therefore, the mean marks are 59.3.
In most of our real life situations, data is usually so large that to make a meaningful study,
it needs to be condensed as grouped data. So, we need to convert ungrouped data into grouped
data and devise some method to find its mean.
Let us convert the ungrouped data of Example 1 into grouped data by forming class-
intervals of width, say 15. Remember that while allocating frequencies to each class-interval,
students falling in any upper class-limit would be considered in the next class, e.g., 4 students
who have obtained 40 marks would be considered in the class-interval 40-55 and not in 25-40.
With this convention in our mind, let us form a grouped frequency distribution table.
Class interval10-25 25-40 40-5555-7070-85 85-100
Number of students2 3 7 6 6 6

333. Mathematic3 2 4
Free Distribution by A.P. Government
Now, for each class-interval, we require a point which would serve as the representative
of the whole class. It is assumed that the frequency of each class-interval is centred
around its mid-point. So, the mid-point of each class can be chosen to represent the
observations falling in the class and is called the class mark. Recall that we find the class mark by
finding the average of the upper and lower limit of the class.
Upper class limit + Lower class limit
Class mark =
2
For the class 10 -25, the class mark is
10 25
2

=17.5. Similarly, we can find the class
marks of the remaining class intervals. We put them in the table. These class marks serve as our
x
i
’s. We can now proceed to compute the mean in the same manner as in the previous example.
Table 14.3
Class Number of Class f
i
x
i
interval students ( f
i
) Marks (x
i
)
10-25 2 17.5 35.0
25-40 3 32.5 97.5
40-55 7 47.5 332.5
55-70 6 62.5 375.0
70-85 6 77.5 465.0
85-100 6 92.5 555.0
Total
i
f=30
i i
f x =1860
The sum of the values in the last column gives us
i i
f x . So, the mean x of the given
data is given by
1860
62
30
i i
i
f x
x
f

  

This new method of finding the mean is known as the Direct Method.
We observe that in the above cases we are using the same data and employing the same
formula for calculating the mean but the results obtained are different. In example 1, 59.3 is the
exact mean and 62 is the approximate mean. Can you think why this is so?

334. Statistics3 2 5
Free Distribution by A.P. Government
THINK AND DISCUSS
1. The mean value can be calculated from both ungrouped and grouped data. Which one
do you think is more accurate? Why?
2. When is it more convenient use grouped data for analysis?
Sometimes when the numerical values of x
1
and f
1
are large, finding the product of x
1
and
f
1
becomes tedious and time consuming. So, for such situations, let us think of a method of
reducing these calculations.
We can do nothing with the f
i
's, but we can change each x
i
to a smaller number so that
our calculations become easy. How do we do this? What about subtracting a fixed number from
each of these x
i
's? Let us try this method for the data in example 1.
The first step is to choose one among the x
i
's as the assumed mean, and denote it by 'a'.
Also, to further reduce our calculation work, we may take 'a' to be that x
i
which lies in the centre
of x
1
, x
2
, ..., x
n
. So, we can choose a = 47.5 or a = 62.5. Let us choose a = 47.5.
The second step is to find the difference between a and each of the x
i
's, that is, the
deviation of ‘a’ from each of the x
i
's, which we denote as d
i
i.e.,d
i
= x
i
– a = x
i
– 47.5
The third step is to find the product of d
i
with the corresponding f
i
, and take the sum of
all the f
i
d
i
’s. These calculations are shown in table 14.4 below-
Table 14.4
Class Number of Class 47.5
i i
d x 
i i
f d
interval students ( f
i
)Marks (x
i
)
10-25 2 17.5 -30 -60
25-40 3 32.5 -15 -45
40-55 7 47.5 0 0
55-70 6 62.5 15 90
70-85 6 77.5 30 180
85-100 6 92.5 45 270
Total
i
f=30
i i
f d = 435
So, from the above table, the mean of the deviations,
i i
i
f d
d
f




335. Mathematic3 2 6
Free Distribution by A.P. Government
Now, let us find the relation between d and x.
Since, in obtaining d
i
we subtracted ‘a’ from each x
i
so, in order to get the mean x we
need to add ‘a’ to d. This can be explained mathematically as:
Mean of deviations,
i i
i
f d
d
f



So,
( )
i i
i
f x a
d
f
 


=
i i i
i i
f x f a
f f
 

 
=
i
i
f
x a
f



= x a
d= x a
Therefore
i i
i
f d
x a
f

 

Substituting the values of a,
i i
f d and
i
f from the table, we get
435
47.5 47.5 14.5 62
30
x    
Therefore, the mean of the marks obtained by the students is 62.
The method discussed above is called the Assumed Mean Method.
ACTIVITY
Consider the data given in example 1 and calculate the arithmetic mean by deviation
method by taking successive values of x
i
i.e., 17.5, 32.5, ... as assumed means. Now discuss
the following:
1. Are the values of arithmetic mean in all the above cases equal?
2. If we take the actual mean as the assumed mean, how much will
i i
f d be?
3. Reason about taking any mid-value (class mark) as assumed mean?

336. Statistics3 2 7
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Observe that in the table given below the values in Column 4 are all multiples of 15. So,
if we divide all the values of Column 4 by 15, we would get smaller numbers which we then
multiply with f
i
. (Here, 15 is the class size of each class interval.)
So, let
i
i
x a
u
h

 where a is the assumed mean and h is the class size.
Now, we calculate u
i
in this way and continue as before (i. e., find f
i
u
i
and then  f
i
u
i
).
Taking h = 15, let us form the table 14.5.
Let
i i
i
i
f u
u
f



Table 14.5
Class Number of Class
i i
d x a 
i
i
x a
u
h


i i
f u
interval students ( f
i
)Marks (x
i
)
10-25 2 17.5 -30 -2 -4
25-40 3 32.5 -15 -1 -3
40-55 7 47.5 0 0 0
55-70 6 62.5 15 1 6
70-85 6 77.5 30 2 12
85-100 6 92.5 45 3 18
Total
i
f=30
i i
f u =29
Here, again let us find the relation between u and x.
We have
i
i
x a
u
h


So
i i
i
f u
u
f



So
( )
i
i
i
x a
f
h
u
f





337. Mathematic3 2 8
Free Distribution by A.P. Government
=
1
i i i
i i
f x f a
h f f
  
 
  
 
=
1
x a
h
 

  
or hu x a 
x a hu 
Therefore,
i i
i
f u
x a h
f
 
  
 
 
or
i i
i
f u
x a h
f
 
   
 
 
Substituting the values of a,
i i
f u and
i
ffrom the table, we get
29
47.5 15
30
x  
= 47.5 + 14.5 = 62
So, the mean marks obtained by a student are 62.
The method discussed above is called the Step-deviation method.
We note that:
· The step-deviation method will be convenient to apply if all the d
i
’s have a common
factor.
· The mean obtained by all the three methods is the same.
· The assumed mean method and step-deviation method are just simplified forms of the
direct method.
· The formulax a hu  still holds if a and h are not as given above, but are any non-
zero numbers such that
i
i
x a
u
h



338. Statistics3 2 9
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Let us apply these methods in other examples.
Example-2. The table below gives the percentage distribution of female teachers in the primary
schools of rural areas of various states and union territories (U.T.) of India. Find the mean
percentage of female teachers using all the three methods.
Percentage of female teachers15 - 2525 - 3535 – 4545 - 5555 - 6565 - 7575 – 85
Number of States/U.T. 6 11 7 4 4 2 1
Source : Seventh All India School Education Survey conducted by NCERT
Solution : Let us find the class marks x
i
of each class, and put them in a table
Here we take a = 50, h = 10,
then d
i
= x
i
– 50 and u
i
=
50
10
i
x
We now find d
i
and u
i
and put them in the table 14.6 :
Table 14.6
PercentageNumber of
i
x
i
d
i
u
i i
f x
i i
f d
i i
f u
of femaleStates/U.T. 50
i
x
50
10
i
x
teachers
15 – 25 6 20 -30 -3 120 -180-18
25 – 35 11 30 -20 -2 330 -220-22
35 – 45 7 40 -10 -1 280 -70 -7
45 – 55 4 50 0 0 200 0 0
55 – 65 4 60 10 1 240 40 4
65 – 75 2 70 20 2 140 40 4
75 – 85 1 80 30 3 80 30 3
Total 35 1390-360-36
From the table above, we obtain 35, 1390, 360, 36
i i i i i i i
f f x f d f u        .
Using the direct method
1390
39.71
35
i i
i
f x
x
f

  

Using the assumed mean method
360
50 50 10.29 39.71
35
i i
i
f d
x a
f
 
      

Using the step-deviation method
36
50 10 39.71
35
i i
i
f u
x a h
f
  
       
 
 
Therefore, the mean percentage of female teachers in the primary schools of rural areas is 39.71.

339. Mathematic3 3 0
Free Distribution by A.P. Government
THINK AND DISCUSS
1.Is the result obtained by all the three methods the same?
2. If x
i
and f
i
are sufficiently small, then which method is an appropriate choice?
3. If x
i
and f
i
are numerically large numbers, then which methods are appropriate to use?
Even if the class sizes are unequal, and x
i
are large numerically, we can still apply the
step-deviation method by taking h to be a suitable divisor of all the d
i
’s.
Example-3. The distribution below shows the number of wickets taken by bowlers in one-day
cricket matches. Find the mean number of wickets by choosing a suitable method. What does
the mean signify?
Number of wickets20 - 6060 - 100100 - 150150 - 250250 - 350 350 – 450
Number of bowlers 7 5 16 12 2 3
Solution : Here, the class size varies, and the x
i
's are large. Let us still apply the step deviation
method with a = 200 and h = 20. Then, we obtain the data as given in the table 14.7 below-
Table 14.7
Number ofNumber of
i
x
i
d
i
u
i i
f u
wickets bowlers ( f
i
) 200
i
x
200
20
i
x
20 – 60 7 40 -160 -8 -56
60 – 100 5 80 -120 -6 -30
100 – 150 16 125-75 -3.75 -60
150 – 250 12 2000 0 0
250 – 350 2 300100 5 10
350 – 450 3 400200 10 30
Total 45 -106
So
106
200 20 200 47.11 152.89
45
i i
i
f u
x a h
f
  
         
 
 
Thus, the average number of wickets taken by these 45 bowlers in one-day cricket is 152.89.

340. Statistics3 3 1
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Classroom Project :
1. Collect the marks obtained by all the students of your class in Mathematics in the recent
examination conducted in your school. Form a grouped frequency distribution of the
data obtained. Do the same regarding other subjects and compare. Find the mean in
each case using a method you find appropriate.
2. Collect the daily maximum temperatures recorded for a period of 30 days in your city.
Present this data as a grouped frequency table. Find the mean of the data using an
appropriate method.
3. Measure the heights of all the students of your class and form a grouped frequency
distribution table of this data. Find the mean of the data using an appropriate method.
EXERCISE - 14.1
1. A survey was conducted by a group of students as a part of their environment awareness
programme, in which they collected the following data regarding the number of plants in
20 houses in a locality. Find the mean number of plants per house.
Number of plants 0 - 22 - 44 – 66 - 8 8 - 1010 - 1212 – 14
Number of houses 1 2 1 5 6 2 3
2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages in ` 100 - 120120 - 140140 - 160160 - 180180– 200
Number of workers 12 14 8 6 10
Find the mean daily wages of the workers of the factory by using an appropriate method.
3. The following distribution shows the daily pocket allowance of children of a locality. The
mean pocket allowance is ` 18. Find the missing frequency f.
Daily pocket 11 - 13 13 - 15 15 - 17 17 - 19 19 - 21 21 - 23 23 - 25
allowance (in ` )
Number of children 7 6 9 13 f 5 4

341. Mathematic3 3 2
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4. Thirty women were examined in a hospital by a doctor and their of heart beats per
minute were recorded and summarised as shown. Find the mean heart beats per minute
for these women, choosing a suitable method.
Number of heart beats/minute65-68 68-71 71-74 74-77 77-80 80-83 83-86
Number of women 2 4 3 8 74 2
5. In a retail market, fruit vendors were selling oranges kept in packing baskets. These
baskets contained varying number of oranges. The following was the distribution of
oranges according to the number of baskets :
Number of oranges10-14 15–19 20-24 25-29 30–34
Number of baskets 15 110 135 115 25
Find the mean number of oranges kept in each basket. Which method of finding the
mean did you choose?
6. The table below shows the daily expenditure on food of 25 households in a locality.
Daily expenditure in ` 100-150 150-200 200-250 250-300 300-350
Number of house holds 4 5 12 2 2
Find the mean daily expenditure on food by a suitable method.
7. To find out the concentration of SO
2
in the air (in parts per million, i.e., ppm), the data
was collected for 30 localities in a certain city and is presented below:
Concentration of SO
2
in ppm 0.00-0.04 0.04-0.08 0.08-0.12 0.12-0.16 0.16-0.20 0.20-0.24
Frequency 4 9 9 2 4 2
Find the mean concentration of SO
2
in the air.
8. A class teacher has the following attendance record of 40 students of a class for the
whole term. Find the mean number of days a student was present out of 56 days in the
term.
Number of days35-38 38-41 41-44 44-47 47-50 50-53 53-56
Number of students 1 3 4 4 7 10 11

342. Statistics3 3 3
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9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.
Literacy rate in %45–55 55-65 65-75 75-85 85-95
Number of cities 3 10 11 8 3
14.3 MODE
A mode is that value among the observations which occurs most frequently.
Before learning about calculating the mode of grouped data let us first recall how we
found the mode for ungrouped data through the following example.
Example-4. The wickets taken by a bowler in 10 cricket matches are as follows: 2 6 4 5 0 2
1 3 2 3. Find the mode of the data.
Solution : Let us arrange the observations in order i.e., 0 1 2 2 2 3 3 4 5 6
Clearly, 2 is the number of wickets taken by the bowler in the maximum number of matches (i.e.,
3). So, the mode of this data is 2.
DO THIS
1. Find the mode of the following data.
a) 5, 6, 9, 10, 6, 12, 3, 6, 11, 10, 4, 6, 7.
b) 20, 3, 7, 13, 3, 4, 6, 7, 19, 15, 7, 18, 3.
c) 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6.
2. Is the mode always at the centre of the data?
3. If another observation is added to the data in Example 4, then does the mode change?
Comment.
4. If the maximum value of an observation in the data in Example 4 is changed to 8 then
would the mode of the data be affected? Comment.
In a grouped frequency distribution, it is not possible to determine the mode by looking
at the frequencies. Here, we can only locate a class with the maximum frequency, called the
modal class. The mode is a value inside the modal class, and is given by the formula.

343. Mathematic3 3 4
Free Distribution by A.P. Government
Mode
1 0
1 0 2
2
f f
l h
f f f
  
      
where, l = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
f
1
= frequency of the modal class,
f
0
= frequency of the class preceding the modal class,
f
2
= frequency of the class succeeding the modal class.
Let us consider the following examples to illustrate the use of this formula.
Example-5. A survey conducted on 20 households in a locality by a group of students resulted
in the following frequency table for the number of family members in a household.
Family size 1-33-55-77-99-11
Number of families 7 8 2 2 1
Find the mode of this data.
Solution : Here the maximum class frequency is 8, and the class corresponding to this frequency
is 3-5. So, the modal class is 3-5.
Now,
modal class = 3-5, lower limit (l) of modal class = 3, class size (h) = 2
frequency of the modal class (f
1
) = 8,
frequency of class preceding the modal class (f
0
) = 7,
frequency of class succeeding the modal class (f
2
) = 2.
Now, let us substitute these values in the formula-
Mode
1 0
1 0 2
2
f f
l h
f f f
  
      
8 7 2
3 2 3 3.286
2 8 7 2 7
 
          
Therefore, the mode of the data above is 3.286.

344. Statistics3 3 5
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Example-6. The marks distribution of 30 students in a mathematics examination are given in the
adjacent table. Find the mode of this data. Also compare and interpret the mode and the mean.
Table 14.8
Class interval Number of Class Marks (x
i
) f
i
x
i
students ( f
i
)
10-25 2 17.5 35.0
25-40 3 32.5 97.5
40-55 7 47.5 332.5
55-70 6 62.5 375.0
70-85 6 77.5 465.0
85-100 6 92.5 555.0
Total
i
f=30
i i
f x =1860
Solution : Since the maximum number of students (i.e., 7) have got marks in the interval, 40-65
the modal class is 40 - 55.
The lower limit ( l ) of the modal class = 40,
The class size ( h) = 15,
The frequency of modal class ( f
1
) = 7,
the frequency of the class preceding the modal class ( f
0
) = 3,
the frequency of the class succeeding the modal class ( f
2
) = 6.
Now, using the formula:
Mode
1 0
1 0 2
2
f f
l h
f f f
  
      
7 3
40 15 40 12 52
2 7 6 3
 
          
Interpretation : The mode marks is 52. Now, from Example 1, we know that the mean marks
is 62. So, the maximum number of students obtained 52 marks, while on an average a student
obtained 62 marks.

345. Mathematic3 3 6
Free Distribution by A.P. Government
THINK AND DISCUSS
1. It depends upon the demand of the situation whether we are interested in finding the
average marks obtained by the students or the average of the marks obtained by most of
the students.
a. What do we find in the first situation?
b. What do we find in the second situation?
2. Can mode be calculated for grouped data with unequal class sizes?
EXERCISE - 14.2
1. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years) 5-15 15-2525-3535-4545-5555-65
Number of patients 6 11 21 23 14 5
Find the mode and the mean of the data given above. Compare and interpret the two
measures of central tendency.
2. The following data gives the information on the observed life times (in hours) of 225
electrical components :
Lifetimes (in hours) 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120
Frequency 10 35 52 61 38 29
Determine the modal lifetimes of the components.
3. The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families. Also, find the
mean monthly expenditure :
Expenditure 1000-1500-2000-2500-3000-3500-4000-4500-
15002000250030003500400045005000
Number of families 24 40 33 28 30 22 16 7

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4. The following distribution gives the state-wise, teacher-student ratio in higher secondary
schools of India. Find the mode and mean of this data. Interpret the two measures.
Number of students15-2020-2525-3030-3535-4040-4545-5050-55
Number of States 3 8 9 10 3 0 0 2
5. The given distribution shows the number of runs scored by some top batsmen of the
world in one-day international cricket matches.
Runs 3000-4000-5000-6000-7000-8000-9000-10000-
400050006000700080009000100001100
Number of batsmen 4 18 9 7 6 3 1 1
Find the mode of the data.
6. A student noted the number of cars passing through a spot on a road for 100 periods,
each of 3 minutes, and summarised this in the table given below.
Number of cars 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80
Frequency 7 14 1312 20 11 158
Find the mode of the data.
14.4 MEDIAN OF GROUPED DATA
Median is a measure of central tendency which gives the value of the middle-most
observation in the data. Recall that for finding the median of ungrouped data, we first arrange the
data values or the observations in ascending order.
Then, if n is odd, the median is the
1
2
th
n 
 
 
 
observation and
if n is even, then the median will be the average of the
2
th
n 
 
 
 
and
1
2
th
n 
 
 
 
observations.

347. Mathematic3 3 8
Free Distribution by A.P. Government
Suppose, we have to find the median of the following data, which gives the marks, out of 50
obtained by 100 students in a test :
Marks obtained 20 29 28 33 42 38 43 25
Number of students6 28 24 15 2 4 1 20
First, we arrange the marks in ascending order and prepare a frequency table 14.9 as follows :
Table 14.9
Marks obtained Number of students (frequency)
20 6
25 20
28 24
29 28
33 15
38 4
42 2
43 1
Total 100
Here n = 100, which is even. The median will be the average of the
2
th
n
and the
1
2
th
n 

  
observations, i.e., the 50
th
and 51
st
observations. To find these observations, we proceed as in
the table 14.9 on above :
Marks obtained Number of studentsCumulative frequency
20 6 6
upto 25 6 + 20 = 26 26
upto 28 26 + 24 = 50 50
upto 29 50 + 28 = 78 78
upto 33 78 + 15 = 93 93
upto 38 93 + 4 = 97 97
upto 42 97 + 2 = 99 99
upto 43 99 + 1 = 100 100

348. Statistics3 3 9
Free Distribution by A.P. Government
Now we add another column depicting this information to the frequency table above and name it
as cumulative frequency column.
From the table above, we see that :
50
th
observation is 28 (Why?)
51
st
observation is 29
Median =
28 29
28.5
2


Remark : Column 1 and column 3 in the above
table are known as Cumulative Frequency Table.
The median marks 28.5 conveys the information that
about 50% students obtained marks less than 28.5
and another 50% students obtained marks more than
28.5.
Consider a grouped frequency distribution
of marks obtained, out of 100, by 53 students, in a
certain examination, as shown in table 14.10.
From the table, try to answer the following questions :
How many students have scored marks
less than 10? The answer is clearly 5.
How many students have scored less than
20 marks? Observe that the number of
students who have scored less than 20
include the number of students who have
scored marks from 0-10 as well as the
number of students who have scored
marks from 10-20. So, the total number
of students with marks less than 20 is 5 +
3, i.e., 8. We say that the cumulative
frequency of the class 10-20 is 8. (As
shown in table 14.11)
Similarly, we can compute the cumulative
frequencies of the other classes, i.e., the
number of students with marks less than
30, less than 40, ..., less than 100.
Table 14.10
Marks Number of students
0-10 5
10-20 3
20-30 4
30-40 3
40-50 3
50-60 4
60-70 7
70-80 9
80-90 7
90-100 8
Table 14.11
Marks obtained Number of students
(Cumulative frequency)
Less than 10 5
Less than 20 5 + 3 = 8
Less than 30 8 + 4 = 12
Less than 40 12 + 3 = 15
Less than 50 15 + 3 = 18
Less than 60 18 + 4 = 22
Less than 70 22 + 7 = 29
Less than 80 29 + 9 = 38
Less than 90 38 + 7 = 45
Less than 100 45 + 8 = 53

349. Mathematic3 4 0
Free Distribution by A.P. Government
This distribution is called the cumulative frequency distribution of the less than type. Here 10, 20,
30, ..., 100, are the upper limits of the respective class intervals.
We can similarly make the table
for the number of students with
scores, more than or equal to 0,
more than or equal to 10, more
than or equal to 20, and so on.
We observe that all 53 students
have scored marks more than or
equal to 0. Since there are 5
students scoring marks in the
interval 0-10, this means that there
are 53-5 = 48 students getting
more than or equal to 10 marks.
Continuing in the same manner,
we get the number of students
scoring 20 or above as 48-3 =
45, 30 or above as 45-4 = 41,
and so on, as shown in the table
14.12.
This table above is called a cumulative frequency distribution of the more than type. Here 0, 10,
20, ..., 90 give the lower limits of the respective class intervals.
Now, to find the median of grouped data, we can make use of any of these cumulative frequency
distributions.
Now in a grouped data, we may not be able to find the middle observation by looking at the
cumulative frequencies as the middle observation will be some value in a class interval. It is,
therefore, necessary to find the value inside a class that divides the whole distribution into two
halves. But which class should this be?
To find this class, we find the cumulative frequencies of all the clases and
2
n
. We now locate the
class whose cumulative frequency is greater than (and nearest to)
2
n
. This is called the median
class.
Table 14.12
Marks obtained Number of students
(Cumulative frequency)
More than or equal to 0 53
More than or equal to 10 53 - 5 = 48
More than or equal to 20 48 - 3 = 45
More than or equal to 30 45 - 4 = 41
More than or equal to 40 41 - 3 = 38
More than or equal to 50 38 - 3 = 35
More than or equal to 60 35 - 4 = 31
More than or equal to 70 31 - 7 = 24
More than or equal to 80 24 - 9 = 15
More than or equal to 90 15 - 7 = 8

350. Statistics3 4 1
Free Distribution by A.P. Government
Table 14.3
Marks Number of students (f) Cumulative frequency (cf)
0-10 5 5
10-20 3 8
20-30 4 12
30-40 3 15
40-50 3 18
50-60 4 22
60-70 7 29
70-80 9 38
80-90 7 45
90-100 8 53
In the distribution above, n = 53. So
2
n
= 26.5. Now 60-70 is the class whose cumulative
frequency 29 is greater than (and nearest to)
2
n
, i.e., 26.5.
Therefore, 60-70 is the median class.
After finding the median class, we use the following formula for calculating the median.
Median
2
n
cf
l h
f
 
     
 
 
 

 
wherel = lower limit of median clas,
n = number of observations,
cf = cumulative frequency of class preceding the median class,
f = frequency of median class,
h = class size (assuming class size to be equal).
Substituting the values
2
n
= 26.5, l = 60, cf = 22, f = 7,h = 10
in the formula above, we get
Median
2
60 10
n
cf
f
 
     
 
 
 

 

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45
60
7
 
66.4
So, about half the students have scored marks less than 66.4, and the other half have scored
marks more than 66.4.
Example-7. A survey regarding the heights (in
cm) of 51 girls of Class X of a school was
conducted and data was obtained as shown in
table. Find their median height.
Solution : To calculate the median
height, we need to find the class
intervals and their corresponding
frequencies. The given distribution
being of the less than type, 140,
145, 150, . . ., 165 give the upper
limits of the corresponding class
intervals. So, the classes should be
below 140, 140 - 145, 145 - 150, .
. ., 160 - 165. (As shown in
table 14.14)
Observe that from the given distribution, we find that there are 4 girls with height less than 140,
i.e., the frequency of class interval below 140 is 4 . Now, there are 11 girls with heights less than
145 an d 4 girls with height less than 140. Therefore, the number of girls with height in the interval
140 - 145 is 11 – 4 = 7. Similarly, the frequencies can be calculated as shown in table. Number
of observations, n = 51
2
n
=
51
2
= 25.5
th
observation which lies in the class 145 - 150.
Height (in cm) Number of girls
Less than 140 4
Less than 145 11
Less than 150 29
Less than 155 40
Less than 160 46
Less than 165 51
Class intervals Frequency Cumulative
frequency
Below 140 4 4
140-145 7 11
145-150 18 29
150-155 11 40
155-160 6 46
160-165 5 51
Table 14.4

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Then, l (the lower limit) = 145,
cf (the cumulative frequency of the class preceding 145 - 150) = 11,
f (the frequency of the median class 145 - 150) = 18,
h (the class size) = 5.
Using the formula, Median
2
n
cf
l h
f
 
  
  
 25.5 11
145 5
18

  
72.5
145
18
  = 149.03
So, the median height of the girls is 149.03 cm. This means that the height of about 50% of the
girls is less than this height, and 50% are taller than this height.
Example-8. The median of the following data is 525. Find the values of x and y, if the total
frequency is 100. Here, CI stands for class interval and Fr for frequency.
CI0-100100-200-300-400-500-600-700-800- 900-
200300400500600700800900 1000
Fr 2 5 X 12 17 20 Y 9 74
Solution :
It is given that n = 100
So, 76 + x + y = 100, i.e., x + y = 24 (1)
The median is 525, which lies in the class 500 – 600
So, l = 500, f = 20, cf = 36 + x, h = 100
Using the formula
Median
2
n
cf
l h
f
 
  
  

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525
50 36
500 100
20
x 
  
i.e., 525 – 500 = (14 – x) × 5
i.e., 25 = 70 – 5x
i.e., 5x = 70 – 25 = 45
So, x = 9
Therefore, from (1), we get 9 + y = 24
i.e., y = 15
Table 14.15
Class intervals FrequencyCumulative frequency
0-100 2 2
100-200 5 7
200-300 X 7+X
300-400 12 19+X
400-500 17 36+X
500-600 20 56+X
600-700 Y 56+X+Y
700-800 9 65+X+Y
800-900 7 72+X+Y
900-1000 4 76+X+Y
We note that:
The median of grouped data with unequal class sizes can also be calculated. However, we shall
not discuss it here.
Now that you have studied about all the three measures of central tendency, let us discuss which
measure would be best suited for a particular requirement.
The mean is the most frequently used measure of central tendency because it takes into account
all the observations, and lies between the extremes, i.e., the largest and the smallest observations
of the entire data. It also enables us to compare two or more distributions. For example, by
comparing the average (mean) results of students of different schools of a particular examination,
we can conclude which school has a better performance.

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However, extreme values in the data affect the mean. For example, the mean of classes having
frequencies more or less the same is a good representative of the data. But, if one class has
frequency, say 2, and the five others have frequency 20, 25, 20, 21, 18, then the mean will
certainly not reflect the way the data behaves. So, in such cases, the mean is not a good
representative of the data.
In problems where individual observations are not important, and we wish to find out a ‘typical’
observation, the median is more appropriate, e.g., finding the typical productivity rate of workers,
average wage in a country, etc. These are situations where extreme values may be there. So,
rather than the mean, we take the median as a better measure of central tendency.
In situations which require establishing the most frequent value or most popular item, the mode is
the best choice, e.g., to find the most popular T.V. programme being watched, the consumer item
in greatest demand, the colour of the vehicle used by most of the people, etc.
EXERCISE - 14.3
1. The following frequency distribution gives the monthly consumption of electricity of 68
consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption65-85 85-105 105-125 125-145 145-165 165-185 185-205
Number of consumers 4 5 1320 14 8 4
2. If the median of 60 observations, given below is 28.5, find the values of x and y.
Class interval0-10 10-20 20-30 30-40 40-50 50-60
Frequency 5 x 20 15 y 5
3. A life insurance agent found the following data for distribution of ages of 100 policy
holders. Calculate the median age, if policies are given only to persons having age 18
years onwards but less than 60 years.
Age Below BelowBelowBelowBelowBelowBelowBelowBelow
(in years) 20 25 30 35 40 45 50 55 60
Number of 2 6 24 45 78 89 92 98 100
policy holders

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4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and
the data obtained is represented in the following table :
Length (in mm)118-126127-135136-144145-153154-162163-171172-180
Number of leaves 3 5 9 12 5 4 2
Find the median length of the leaves. (Hint : The data needs to be converted to continuous
classes for finding the median, since the formula assumes continuous classes. The classes then
change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)
5. The following table gives the distribution of the life-time of 400 neon lamps
Life time1500-2000-2500-3000-3500-4000-4500-
(in hours)2000250030003500400045005000
Number of14 56 60 86 74 62 48
lamps
Find the median life time of a lamp.
6. 100 surnames were randomly picked up from a local telephone directory and the frequency
distribution of the number of letters in the English alphabet in the surnames was obtained
as follows
Number of letters 1-4 4-7 7-1010-1313-1616-19
Number of surnames 6 30 40 16 4 4
Determine the median number of letters in the surnames. Find the mean number of letters in the
surnames? Also, find the modal size of the surnames.
7. The distribution below gives the weights of 30 students of a class. Find the median
weight of the students.
Weight (in kg)40-4545-5050-5555-6060-6565-7070-75
Number of students 2 3 8 6 6 3 2
14.5 GRAPHICAL REPRESENTATION OF CUMULATIVE FREQUENCY DISTRIBUTION
As we all know, pictures speak better than words. A graphical representation helps us in
understanding given data at a glance. In Class IX, we have represented the data through bar
graphs, histograms and frequency polygons. Let us now represent a cumulative frequency
distribution graphically.

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For example, let us consider the cumulative frequency distribution given in example?
Recall that the values 10, 20, 30, ..., 100 are the upper limits of the respective class intervals. To
represent the data in the
table graphically, we
mark the upper limits of
the class intervals on the
horizontal axis (X-axis)
and their corresponding
cumulative frequencies
on the vartical axis
(Y-axis), choosing a
convenient scale. The
scale may not be the
same on both the axis. Let us now plot the points corresponding to the ordered pairs given by
(upper limit, corresponding cumulative frequency), i.e., (10, 5), (20, 8), (30,, 12), (40, 15), (50,
18), (60, 22), (70, 29), (80, 38), (90, 45), (100, 53) on a graph paper and join them by a free
hand smooth curve. The curve we get is called a cumulative frequency curve, or an ogive (of the
less than type). (See in above figure)
The term 'ogive is pronounced as 'ojeev' and is derived from the word ogee. An ogee is
a shape consisting of a concave arc flowing into a convex arc, so forming an S-shaped
curve with vertical ends. In architecture, the ogee shape is one of the characteristics of
the 14
th
and 15
th
century Gothic styles.
Next, again we consider the cumulative frequency distribution and draw its ogive (of the more
than type).
Recall that, here 0, 10, 20, ...., 90 are the
lower limits of the respective class intervals
0-10, 10-20, ....., 90-100. To represent 'the
more than type' graphically, we plot the lower
limits on the X-axis and the corresponding
cumulative frequencies on the Y-axis. Then we
plot the points (lower limit, corresponding
cumulative frequency), i.e., (0, 53), (10, 48),
(20, (45), (30, 41), (40, 38), (50, 35),
(60, 31), (70, 24), (80, 15), (90, 8), on a
graph paper, and join them by a free hand
0
10
20
30
40
50
60
0 10 20 30 40 50 60 70 80 90100
Cumulative frequency
Upper limits
'Less than' ogive
0
10
20
30
40
50
60
0102030405060708090
Cumulative frequency
Lower limits
'More than' ogive

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smooth curve. The curve we get is a cumulative frequency curve, or an ogive (of the more than
type). (See the last figure).
Remark : Note that both the ogives (the last page figures) correspond to the same data, which
is given in Table 14.11 and 14.12.
Now are the ogives related to the median in any way? Is it possible to obtain the median from
these two cumulative frequency curves corresponding to the data in Table 14.12? Let us see.
One obvious way is to locate on
53
26.5
2 2
n
  on the y-axis (see
Fig. -----). From this point, draw a
line parallel to the x-axis cutting the
curve at a point. From this point,
draw a perpendicular to the x-axis.
The point of intersection of this
perpendicular with the x-axis
determines the median of the data
(see the adjacent figures).
Another way of obtaining the median is the following :
Draw both ogives (i.e., of the less than type and of the more than type) on the same axis. The two
ogives will intersect each other at a point. From this point, if we draw a perpendicular on the x-
axis, the point at which it cuts the x-axis gives us the median (see the figure below).
0
10
20
30
40
50
60
0 10 20 30 40 50 60 70 80 90100
Upper limits
C
u
m
u
l
a
ti
v
e

f
r
e
q
u
e
n
c
y
Median (66.4)
0
10
20
30
40
50
60
0 10 20 30 40 50 60 70 80 90 100
Cumulative frequency
Median (66.4)

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Example-9. The annual profits earned by 30 shops in a locality give rise to the following
distribution :
Profit (in lakhs ` ) Number of shops (frequency)
More than or equal to 5 30
More than or equal to 10 28
More than or equal to 15 16
More than or equal to 20 14
More than or equal to 25 10
More than or equal to 30 7
More than or equal to 35 3
Draw both ogives for the data above. Hence obtain the median profit.
Solution : We first draw the coordinate axes, with lower limits of the profit along the horizontal
axis, and the cumulative frequency along the vertical axes. Then, we plot the points (5, 30), (10,
28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3). We join these points with a smooth curve
to get the more than ovive, as shown in the figure below-
0
5
10
15
20
25
30
35
0 5 10 15 20 25 30 35
Lower limits of profit (in lakhs Rs.)
Cumulative frequency

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0
5
10
15
20
25
30
35
0 5 10 15 20 25 30 35 40
Cumulative frequency
Median (17.5)
Profit (in lakhs Rs.)
Now, let us obtain the classes, their frequencies and the cumulative frequency from the table
above.
Table 14.16
Classes 5-1010-1515-2020-2525-3030-35 35-40
Number of shops 2 12 2 4 34 3
Cumulative frequency2 14 16 20 2327 30
Using these values, we plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27),
(40, 30) on the same axes as in last figure to get the less than ogive, as shown in figure below.
The abcissa of their point of intersection is nearly 17.5, which is the median. This can also be
verified by using the formula. Hence, the median profit (in lakhs) is ` 17.5.
Remark : In the above examples, it may be noted that the class intervals were continuous. For
drawing ogives, it should be ensured that the class intervals are continuous.
EXERCISE - 14.4
1. The following distribution gives the daily income of 50 workers of a factory.
Daily income (in `)100-120 120-140 140-160 160-180 180-200
Number of workers 12 14 8 6 10
Convert the distribution above to a less than type cumulative frequency distribution, and draw its
ogive.

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2. During the medical check-up of 35 students of a class, their weights were recorded as
follows :
Weight (in kg) Number of students
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph
and verify the result by using the formula.
3. The following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield50-5555-6060-65 65-70 70-75 75-80
(in kg/ha)
Number of farms 2 8 12 24 38 16
Change the distribution to a more than type distribution, and draw its ogive.
WHAT WE HAVE DISCUSSED
In this chapter, you have studied the following points :
1. The mean for grouped data can be found by :
(i)The direct method :
i i
i
f x
x
f



(ii)The assumed mean method :
i i
i
f d
x a
f

 

(iii)The step deviation method :
i i
i
f u
x a h
f
  
     

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Here, the assumption is that the frequency of a class is centred at its mid-point, called its
class mark.
2. The mode for grouped data can be found by using the formula :
Mode
1 0
1 0 2
2
f f
l h
f f f
  
      
where, symbols have their usual meaning.
3. The cumulative frequency of a class is the frequency obtained by adding the frequencies
of all the classes preceding the given class.
4. The median for grouped data is formed by using the formula :
Median
2
n
cf
l h
f
 
     
 
 
 

 
Where symbols have their usual meanings.
5. Representing a cumulative frequency distribution graphically as a cumulative frequency
curve, or an ogive of the less than type and of the more than type.
6. The median of grouped data can be obtained graphically as the x-coordinate of the point
of intersection of the two ogives for this data.

363. A.I.1 INTRODUCTION
On 25
th
February 2013, the ISRO launcher PSLV C20, put the satellite SARAL into
orbit. The satellite weighs 407 kg. It is at an altitude of 781 km and its orbit is inclined at an angle
of 98.5º.
On reading the above information, we may wonder:
(i)How did the scientists calculate the altitude as 781km. Did they go to space and measure
it?
(ii)How did they conclude that the angle of orbit is 98.5º

without actually measuring?
Some more examples are there in our daily life where we wonder how the scientists and
mathematicians could possibly have estimated these results. Observe these examples:
(i)The temperature at the surface of the sun is about 6,000ºC.
(ii)The human heart pumps 5 to 6 liters of blood in the body every 60 Seconds.
(iii)We know that the distance between the sun and the earth is 1,49,000 km.
In the above examples, we know that no one went to the sun to measure the temperature
or the distance from earth. Nor can we take the heart out of the body and measure the blood it
pumps. The way we answer these and other similar questions is throughm a t h e m a t i c a l
modelling.
Mathematical modelling is used not only by scientists but also by us. For example, we
might want to know how much money we will get after one year if we invest Rs. 100 at 10%
simple interest. Or we might want to know how many litres of paint is needed to whitewash a
room. Even these problems are solved by mathematical modelling.
THINK AND DISCUSS
Discuss with your friends some more examples in real life where we cannot directly
measure and must use mathematical modelling .
Mathematical ModellingI

364. Mathematic3 5 4
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A.I.2 MATHEMATICAL MODELS
Do you remember the formula to calculate the area of a triangle?
Area of Triangle =
1
2
 base  height.
Similarly, simple interest calculation uses the formula
100
PTR
I . This formula or equation
is a relation between the Interest (I); Principle (P); Time (T); and Rate of Interest (R). These
formulae are examples of mathematical models.
Some more examples for mathematical models.
(i)Speed (S)
tan ( )
( )
Dis ce d
time t

(ii)In compound interest sum (A) = 1
100
n
r
P
 
  
Where P = Principle
r = rate of interest
n = no.of years
So, Mathematical model is nothing but a mathematical description or relation that
describes some real life situation.
DO THIS
Write some more mathematical models which you have learnt in previous classes.
A.I.3 MATHEMATICAL MODELLING
We often face problems in our day to day life. To solve them, we try to write it as an
equivalent mathematical problem and find its solution. Next we interpret the solution and check
to what extent the solution is valid. This process of constructing a mathematical model and using
it to find the answer is known as mathematical modelling.

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Example-1. Vani wants to buy a TV that costs Rs. 19,000 but she has only Rs. 15,000. So she
decides to invest her money at 8% simple interest per year. After how many years will she be
able to buy the TV?
Step 1 : (Understanding the problem): In this stage, we define the real problem. Here, we are
given the principal, the rate of simple interest and we want to find out the number of years after
which the amount will become Rs. 19000.
Step 2 : (Mathematical description and formulation) In this step, we describe, in mathematical
terms, the different aspects of the problem. We define variables, write equations or inequalities
and gather data if required.
Here, we use the formula for simple interest which is

100
PTR
I (Model) where
P = PrincipleT = number of yearsR = rate of interest I = Interest
We need to find time =
100I
T
RP

Step 3: (Solving the mathematical problem) In this step, we solve the problem using the
formula which we have developed in step 2.
We know that Vani already has Rs. 15,000 which is the principal, P
The final amount is Rs. 19000 so she needs an extra (19000-15000) = 4000. This will
come from the interest, I .
Rate = 8%, then I = 4000 =
150 00 8
10
T 
0
or Step4 : (Interpreting the solution): The solution obtained in the previous step is interpretated
here.
Here T = 3
1
3
. This means three and one third of a year or three years and 4 months.
So, Vani can buy a washing machine after 3 year 4 months
T =
4000
1200
= 3
1
3
years

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Step5 : (Validating the model): We can’t always accept a model that gives us an answer that
does not match the reality. The process of checking and modifying the mathematical model, if
necessary, is validation.
In the given example, we are assuming that the rate of interest will not change. If the rate
changes then our model
100
PTR
will not work. We are also assuming that the price of the washing
machine will remain Rs. 19,000.
Let us take another example.
Example-2. A class has 50 children and 1 teacher. They want to go from Warangal to Hyderabad
by jeeps. Each jeep can hold six persons not including driver. How many jeeps are needed?
Step 1 : We want to find the number of jeeps needed to carry 51 persons, given that each jeep
can seat 6 persons besides the driver.
Step 2 : Number of jeeps = (Number of persons) / (Persons that can be seated in one jeep)
Step 3 : Number of jeeps = 51/6 = 8.5
Step 4 : Interpretation
We know that it is not possible to have 8.5 jeeps. So, the number of jeeps needed has to
be the nearest whole number which is 9.
Step 5 : Validation
While modelling, we have assumed that adults and children occupy same space.
DO THIS
1. Take any word problem from your textbook, make a mathematical model for the chosen
problem and solve it.
2. Make a mathematical model for the problem given below and solve it.
Suppose a car starts from a place A and travels at a speed of 40 Km/n towards another
place B. At the same time another car starts from B and travels towards A at a speed of
30 Km/n. If the distance between A and B is 100 km; after how much time will that cars
meet?
So far, we have made mathematical models for simple word problems. Let us take a real
life example and model it.

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Example-3. In the year 2000, 191 member countries of the U.N.signed a declaration to
promote gender equality. One indicator for deciding whether this goal has been achieved is the
ratio of girls to boys in primary, secondary and tertiary education. India also signed the declaration.
The data for the percentage of girls in India who are enrolled in primary schools is given in Table
A.I.1. Table A.I.1
Year Enrolment (in %)
1991 – 92 41.9
1992 – 93 42.6
1993 – 94 42.7
1994 – 95 42.9
1995 – 96 43.1
1996 – 97 43.2
1997 -98 43.5
1998 – 99 43.5
1999 – 2000 43.6
2000 – 01 43.7
2001 - 02 44.1
Using this data, mathematically describe the rate at which the proportion of girls enrolled
in primary schools grew. Also, estimate the year by which the enrollment of girls will reach 50%.
Solution :
Step 1 : Formulation Let us first convert the problem into a mathematical problem.
Table A2.1 gives the enrollment for the years 1991 – 92, 1992- 93 etc. Since the students
join at the beginning of an academic year, we can take the years as 1991, 1992 etc. Let us
assume that the percentage of girls who join primary schools will continue to grow at the same
rate as the rate in Table A2.1. So, the number of years is important, not the specific years. (To
give a similar situation, when we find the simple interest for say, Rs.15000 at the rate 8% for
three years, it does not matter whether the three – year period is from 1999 to 2002 or from
2001 to 2004. What is important is the interest rate in the years being considered)
Here also, we will see how the enrollment grows after 1991 by comparing the number of
years that has passed after 1991 and the enrollment. Let us take 1991 as the 0
th
year, and write

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1 for 1992 since 1 year has passed in 1992 after 1991. Similarly we will write 2 for 1993, 3 for
1994 etc. So, Table A2.1 will now look like as Table A.I.2
Table A.I.2
Year Enrolment (in%)
0 41.9
1 42.6
2 42.7
3 42.9
4 43.1
5 43.2
6 43.5
7 43.5
8 43.6
9 43.7
10 44.1
The increase in enrolment is given in the following table A.I.3.
Table A.I.3
Year Enrolment(in%) Increase
0 14.9 0
1 42.6 0.7
2 42.7 0.1
3 42.9 0.2
4 43.1 0.2
5 43.2 0.1
6 43.5 0.3
7 43.5 0
8 43.6 0.1
9 43.7 0.1
10 44.1 0.4

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At the end of the one year period from 1991 to 1992, the enrollment has increased by
0.7% from 41.9% to 42.6%. At the end of the second year, this has increased by 0.1% from
42.6% to 42.7%. From the table above, we cannot find a definite relationship between the
number of years and percentage. But the increase is fairly steady. Only in the first year and in the
10
th
year there is a jump. The mean of these values is

0.7 0.1 0.2 0.1 0.1 0 0.1 0.1 0.4
10
       
= 0.22
Let us assume that the enrolment steadily increases at the rate of 0.22 percent.
Step 2 : Mathematical Description
We have assumed that the enrolment increases steadily at the rate of 0.22% per year.
So, the Enrolment Percentage (EP) in the first year = 41.9 + 0.22
EP in the second year = 41.9 + 0.22 + 0.22 = 41.9 + 2 x 0.22
EP in the third year = 41.9 + 0.22 + 0.22 + 0.22 = 41.9 + 3 x 0.22
So, the enrolment percentage in the n
th
year = 41.9 + 0.22n, for n 1.
Now, we also have to find the number of years by which the enrolment will reach 50%.
So, we have to find the value of n in the equation or formula
50 = 41.9 + 0.22n
Step 3 : Solution : Solving (2) for n, we get
N =
50 41.9
0.22

=
8.1
0.22
= 36.8
Step 4 : Interpretation : Since the number of years is an integral value, we will take the next
higher integer, 37. So, the enrolment percentage will reach 50% in 1991 + 37 = 2028.
Step 4 : Validation Since we are dealing with a real life situation, we have to see to what extent
this value matches the real situation.
Let us check Formula (2) is in agreement with the reality. Let us find the values for the
years we already know, using Formula (2), and compare it with the known values by finding the
difference. The values are given in Table A.I.4.

370. Mathematic3 6 0
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Table A.I.4
Year EnrolmentValues given by (2) Difference
(in %) (in %) (in %)
0 41.9 41.90 0
1 42.6 42.12 0.48
2 42.7 42.34 0.36
3 42.9 42.56 0.34
4 43.1 42.78 0.32
5 43.2 43.00 0.20
6 43.5 43.22 0.28
7 43.5 43.44 0.06
8 43.6 43.66 -0.06
9 43.7 43.88 -0.18
10 44.1 44.10 0.00
As you can see, some of the values given by Formula (2) are less than the actual values
by about 0.3% or even by 0.5%. This can give rise to a difference of about 3 to 5 years since the
increase per year is actually 1% to 2%. We may decide that this much of a difference is acceptable
and stop here. In this case, (2) is our mathematical model.
Suppose we decide that this error is quite large, and we have to improve this model.
Then, we have to go back to Step 2, and change the equation. Let us do so.
Step 1 : Reformulation : We still assume that the values increase steadily by 0.22%, but we
will now introduce a correction factor to reduce the error, For this, we find the mean of all the
errors. This is
0 0.48 0.36 0.34 0.2 0.28 0.06 0.06 0.18 0
0.2210
        
= 0.18
We take the mean of the errors, and correct our formula by this value.
Revised Mathematical Description : Let us now add the mean of the errors to our formula
for enrolment percentage given in (2). So, our corrected formula is :

371. Mathematical Modelling3 6 1
Free Distribution by A.P. Government
Enrolment percentage in the nth year
= 41.9 + 0.22n + 0.18 = 42.08 + 0.22n, for n 1 (3)
We will also modify Equation (2) appropriately. The new equation for n is :
50 = 42.08 + 0.22n (4)
Altered Solution : Solving Equation (4) for n, we get
n =
50 42.08
0.22

=
7.92
0.22
= 36
Interpretation : Since n = 36, the enrolment of girls in primary schools will reach 50% in the
year 1991 + 36 = 2027.
Validation : Once again, let us compare the values got by using Formula (4) with the actual
values. Table A.I.5 gives the comparison.
Table A.I.5
YearEnrolmentValues given DifferenceValues Difference
(in %) by (2) between given between
Values by (4) values
0 41.9 41.90 0 41.9 0
1 42.6 42.12 0.48 42.3 0.3
2 42.7 42.34 0.36 42.52 0.18
3 42.9 42.56 0.34 42.74 0.16
4 43.1 42.78 0.32 42.96 0.14
5 43.2 43.00 0.20 43.18 0.02
6 43.5 43.22 0.28 43.4 0.1
7 43.5 43.44 0.06 43.62 -0.12
8 43.6 43.66 -0.06 43.84 -0.24
9 43.7 43.88 -0.18 44.06 -0.36
10 44.1 44.10 0.00 44.28 -0.18
As you can see, many of the values that (4) gives are closer to the actual value than the
values that (2) gives. The mean of the errors is 0 in this case.

372. Mathematic3 6 2
Free Distribution by A.P. Government
A.I.4 ADVANTAGES OF MATHEMATICS MODELING
1. The aim of mathematical modeling is to get some useful information about a real world
problem by converting it into mathematical problem. This is especially useful when it is
not possible or very expensive to get information by other means such as direct observation
or by conducting experiments.
For example, suppose we want to study the corrosive effect of the discharge of the
Mathura refinery on the Taj Mahal. We would not like to carry out experiments on the
Taj Mahal directly because that would damage a valuable monument. Here mathematical
modeling can be of great use.
2. Forecasting is very important in many types of organizations, since predictions of future
events have to be incorporated into the decision – making process.
For example
(i)In marketing departments, reliable forecasts of demand help in planning of the
sale strategies
(ii)A school board needs to able to forecast the increase in the number of school
going children in various districts so as to decide where and when to start new
schools.
3. Often we need to estimate large values like trees in a forest; fishes in a lake; estimation of
votes polled etc.
Some more examples where we use mathematical modelling are:
(i)Estimating future population for certain number of years
(ii)Predicting the arrival of Monsoon
(iii)Estimating the literacy rate in coming years
(iv)Estimating number of leaves in a tree
(v)Finding the depth of oceans
A.I.5 LIMITATIONS OF MATHEMATICAL MODELING
Is mathematical modeling the answer to all our problems?
Certainly not; it has it’s limitations. Thus, we should keep in mind that a model is only a
simplification of a real world problem, and the two are not same. It is some thing like the difference
between a map that gives the physical features of a country, and the country itself. We can find
the height of a place above the sea level from this map, but we cannot find the characteristics of
the people from it. So, we should use a model only for the purpose it is supposed to serve,
remembering all the factors we have neglected while constructing it. We should apply the model
only within the limits where it is applicable.

373. Mathematical Modelling3 6 3
Free Distribution by A.P. Government
A.I.6 TO WHAT EXTENT WE SHOULD TRY TO IMPROVE OUR MODEL?
To improve a model we need to take into account several additional factors. When we
do this we add more variables to our mathematical equations. The equations becomes complicated
and the model is difficult to use. A model must be simple enough to use yet accurate; i.e the closer
it is to reality the better the model is.
TRY THIS
A problem dating back to the early 13
th
century, posed by Leonardo Fibonacci, asks
how many rabbits you would have in one year if you started with just two and let all of them
reproduce. Assume that a pair of rabbits produces a pair of offspring each month and that each
pair of rabbits produces their first offspring at the age of 2 months. Month by month, the number
of pairs of rabbits is given by the sum of the rabbits in the two preceding months, except for the
0
th
and the 1
st
months. The table below shows how the rabbit population keeps increasing every
month.
Month Pairs of Rabbits
0 1
1 1
2 2
3 3
4 5
5 8
6 13
7 21
8 34
9 55
10 89
11 144
12 233
13 377
14 610
15 987
16 1597
After one year, we have 233 rabbits. After just 16 months, we have nearly 1600 pairs of
rabbits.
Clearly state the problem and the different stages of mathematical modelling in this situation.

374. Mathematic3 6 4
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We will finish this chapter by looking at some interesting examples.
Example-4. (Rolling of a pair of dice) : Suppose your teacher challenges you to the following
guessing game : She would throw a pair of dice. Before that you need to guess the sum of the
numbers that show up on the dice. For every correct answer, you get two points and for every
wrong guess you lose two points. What numbers would be the best guess?
Solution :
Step 1 (Understanding the problem) : You need to know a few numbers which have higher
chances of showing up.
Step 2 (Mathematical description) : In mathematical terms, the problem translates to finding
out the probabilities of the various possible sums of numbers that the dice could show.
We can model the situation very simply by representing a roll of the dice as a random
choice of one of the following thirty six pairs of numbers.
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
The first number in each pair represents the number showing on the first die, and the
second number is the number showing on the second die.
Step 3 (Solving the mathematical problem) : Summing the numbers in each pair above, we
find that possible sums are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. We have to find the probability
for each of them, assuming all 36 pairs are equally likely.
We do this in the following table.
Sum 23 4 5 6 7 8 9 101112
Probability
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Observe that the chance of getting a sum of a seven is
1
6
, which is larger than the
chances of getting other numbers as sums.

375. Mathematical Modelling3 6 5
Free Distribution by A.P. Government
Step 4 (Interpreting the solution) : Since the probability of getting the sum 7 is the highest, you
should repeatedly guess the number seven.
Step 5 (Validating the model) : Toss a pair of dice a large number of times and prepare a
relative frequency table. Compare the relative frequencies with the corresponding probabilities.
If these are not close, then possibly the dice are biased. Then, we could obtain data to evaluate
the number towards which the bias is.
Before going to the next try this exercise, we need some background information.
Not having the money you want when you need it, is a common experience for many
people. Whether it is having enough money for buying essentials for daily living, or for buying
comforts, we always require money. To enable the customers with limited funds to purchase
goods like scooters, refrigerators, televisions, cars, etc., a scheme known as an instalment scheme
(or plan) is introduced by traders.
Sometimes a trader introduces an instalment scheme as a marketing strategy to allow
customers to purchase these articles. Under the instalment scheme, the customer is not required
to make full payment of the article at the time of buying it. She/he is allowed to pay a part of it at
the time of purchase and the rest can be paid in instalments, which could be monthly, quarterly,
half-yearly, or even yearly. Of course, the buyer will have to pay more in the instalment plan,
because the seller is going to charge some interest on account of the payment made at a later date
(called deferred payment).
There are some frequently used terms related to this concept. You may be familiar with
them. For example, the cash price of an article is the amount which a customer has to pay as full
payment of the article at the time it is purchased. Cash down payment is the amount which a
customer has to pay as part payment of the price of an article at the time of purchase.
Now, try to solve the problem given below by using mathematical modelling.
TRY THIS
Yashodhara wants to buy a bicycle. She goes to the market and finds that the bicycyle of
her choice costs Rs. 1800. She only has Rs. 600 with her. To help, the shopkeepr offers to help
her. He says that Yashodhara can make a down payment of Rs. 600 and pay the rest in monthly
instalments of Rs. 610 each. Yashodhara can either take the shopkeepers offer or go to a bank
and take a loan at 10 % per annum simple interest. What should she do - take the shopkeeper's
offer or go to the bank?

377. Mathematic3 6 6
Free Distribution by A.P. Government
EXERCISE - 1.1
1. (i)Terminating(ii)Non-terminating(iii)Terminating
(iv)Terminating(v)Non-terminating
2. (i)
3
4
(ii)
1
3
2
(iii)
31
25
3. (i)Rational (ii)Irrational(iii)Rational (iv) Rational
(v)Rational (vi)Irrational(vii)Rational
EXERCISE - 1.2
1. (i)2
2
 5 7(ii)2
2
 3  13(iii)3
2
 5
2
17
(iv)5  7  11 1 13 (v)17  19  23
2. (i)LCM = 420, HCF = 3 (ii)LCM = 11339, HCF = 1
(iii)LCM = 1800, HCF = 1 (iv)LCM = 216, HCF = 36
(v)LCM = 22338, HCF = 9
EXERCISE - 1.3
1. (i)0.375 (terminating)(ii)0.5725 (terminating)(iii) 4.2 (terminating)
(iv)0.18 (non-terminating, repeating)(v)0.064 (terminating)
2. (i)Terminating (ii)Non-terminating,repeating
(iii)Non-terminating,repeating (iv)Terminating
(v)Non-terminating, repeating (vi)Terminating
(vii)Non-terminating, repeating(viii)Terminating(ix)Terminating
(x)Non-terminating, repeating
Answers

378. Answers 3 6 7
Free Distribution by A.P. Government
3. (i)0.52(ii)0.9375 (iii)0.115 (iv)32.08(v)1.3
4. (i)Rational (ii)Not a rational (iii)Rational
5. m = 5, n = 3
6. m = 4, n = 2
EXERCISE - 1.5
1. (i)log
3
243 = 5(ii)log
2
1024 = 10 (iii)log
10
1000000 = 6
(iv)log
10
0.001 = 3 (v)log
3
1
9
= 2 (vi)log
6
1 = 0
(vii)log
5
1
5
= 1(viii)
49
log 7 = 1(ix) log
27
9 =
2
3
(x)log
32
1
4
= 
2
5
2. (i)18
2
= 324 (ii)10
4
= 10000(iii)a
b
= x
iv)4x

= 8 (v)3y =
1
27
3. (i)
1
2
(ii)
1
4
(iii)4( iv)0
(v)
1
2
(vi)9 (vii)2(viii)3
4. (i)log 10 (ii)log 8 (iii)log 64
(iv)log
9
8
(v)log 243 (vi)log 45
5. (i)3(log 2 + log 5) (ii)7log 2  4log 5(iii) 2log x + 3 log y + 4 log z
(iv)2log p + 3 log q log r (v)
1
2
(3 log x 2 log y)
EXERCISE - 2.1
1. (i)Set(ii)Not set (iii)Not set
(iv)Set(v)Set

379. Mathematic3 6 8
Free Distribution by A.P. Government
2. (i) (ii) (iii) (iv)
(v) (vi)
3. (i)x A (ii)B = {d} (iii)1  N (iv)8 P
4. (i)Not true (ii)Not true
(iii)True (iv)Not true
5. (i)B = {1, 2, 3, 4, 5}
(ii)C = {17, 26, 35, 44, 53, 62, 71}
(iii)D = {5, 3}
(iv)E = {B, E, T, R}
6. (i)A = {x : x is multiple of 3 & less than 13}
(ii)B = {x : x is in power of 2
x
& x is less than 6}
(iii)C = {x : x is in power of 5 & x less than 5)
(iv)D = {x : x in square of natural number and not greater than 10}
7. (i)A = {51, 52, 53, .................., 98, 99}
(ii)B = {+2, 2}
(iii)D = {2, 0, 4, A, 2}
8. (i)(c)
(ii)(a)
(iii) (d)
(iv) (b)
EXERCISE - 2.2
1. (i)Not empty(ii)Empty (iii)Empty
(iv)Empty (v)Not empty
2. (i)Finite (ii)Finite (iii)Finite
3. (i)Finite (ii)Infinite (iii)Infinite (iv) Infinite

380. Answers 3 6 9
Free Distribution by A.P. Government
EXERCISE - 2.3
1. Yes, equal set
2. (i)Equal (ii)Not equal(iii)Equal (iv)Not equal
(v)Not equal(vi)Not equal(vii)Not equal
3. (i)A = B (ii)A B (iii)A = B (iv)A B
EXERCISE - 2.4
1. (i)True (ii)Not false(iii)False (iv)False
2. (i)x is greater than 30
(ii)x = 2x + 1 means x is odd
(iii)x is multiple of 15. So 5 does not exist
(iv)x is prime number but 9 is not a prime number
3.(i){p}, {q}, {p, q}, {}
(ii){x}, {y}, {x, y}, {}
(iii){a}, {b}, {c}, {d}, {a, b}, {b, c}, {c, d}, {a, c}, {a, d}, {b, d}, {a, b, c},
{b, c, d}, {a, b, d}, {a, c, d}, {a, b, c, d)
EXERCISE - 2.5
1. Yes, AB & BB are same
2. A   = {0, 2, 4}
A  A = {}
3. A B = {2, 4, 8, 10}
B A = {3, 9, 12, 15}
4. A  B = B
5. A  B = {odd natural number}
{1, 3, 5, ............}
A  C = {}
A  D = {4, 6, 8, 9, 10, 12, ................. 100}
B  C = {odd natural number}
B  D = {9, 15, 21, 25, 27, 33, ......... 99}
C  D = {4, 6, 8, 9, ................ 99}

381. Mathematic3 7 0
Free Distribution by A.P. Government
6. (i)A  B = {3, 6, 9, 15, 18, 21}
(ii)A  C = {3, 9, 15, 18, 21}
(iii)A  D = {3, 6, 9, 12, 18, 21}
(iv)B  A = {4, 8, 16, 20}
(v)C  A = {2, 4, 8, 10, 14, 16}
(vi)D  A = {5, 10, 20}
(vii)B  C = {20}
(viii)B  D = {4, 8, 12, 16}
(ix)C  B = {2, 6, 10, 14)
(x)D  B = {5, 10, 15}
7. (i)False
(ii)True
(iii)True
(iv)True
EXERCISE - 3.1
1. (a)(i)6 (ii)7 (iii)6
(b)Left to children
2. (i)False ( 2 is coefficient of x
2
not a degree)
(ii)False (Coefficient of x
2
is 4)
(iii)True (For any constant term, degree is zero)
(iv)False (It is not a pdynomial at all)
(v)False (Degree of a pohynomial is not related with number of terms)
3. p(1) = 0, p (1) = 2, p(0) = 1, p(2) = 7, p(2) = 9
4. Yes, 2 and 2 are zeroes of the pohynomial x
4
-16
5. Yes, 3 and 2 are zeroes of the pohynomial x
2
x6

382. Answers 3 7 1
Free Distribution by A.P. Government
EXERCISE - 3.2
1. (i)No zeroes(ii)1 (iii)3
(iv)2 (v)4 (vi)3
2. (i)0 (ii)2, 3 (iii)2, 3 (iv) 2, 2, 4 
3. (i)4, 3 (ii)3, 3(iii)No zeroes
(iv)4, 1 (v)1, 1
4. p
1
4
 

 
= 0 and p (1) = 0
EXERCISE - 3.3
1. (i)4, 2 (ii)
1 1
,
2 2
(iii)
3 1
,
2 3

(iv)0, 2 (v) 15 15 (vi)
4
1,
3

2. (i)4x
2
 x 4(ii)
2
3 3 2 1x x  (iii)
2
5x
(iv)
2
1x x  (v)
2
4 1x x  (vi)
2
4 1x x 
3. (i)x
2
 x 2(ii)
2
3x (iii)
2
4 3 1x x 
(iv)
2
4 8 3x x 
4. Yes, all are zeroes
EXERCISE - 3.4
1. (i)Quotient = x 3 and remainder = 7x 9
(ii)Quotient = x
2
+ x 3 and remainder = 8
(iii)Quotient = x
2
2 and remainder = 5x + 10
2. (i)Yes(ii)Yes(iii)No
3. 1, 1
4. g(x) = x
2
x + 1
5. (i)p(x) = 2x
2
2x + 14, g(x) = 2, q(x) = x
2
x + 7, r(x) = 0

383. Mathematic3 7 2
Free Distribution by A.P. Government
(ii)p(x) = x
3
+ x
2
+ x + 1, g(x) = x
2
1, q(x) = x + 1, r(x) = 2x + 2
(iii)p(x) = x
3
+ 2x
2
x +2, g(x) = x
2
1, q(x) = x + 2, r(x) = 4
There can be several examples in each of (i), (ii) and (iii) which you can give.
EXERCISE - 4.1
1. (a)Intersect at a point
(b)Coincident
(c)Parallel
2. (a)Consistent(b)Inconsistent (c)Consistent
(d)Consistent(e)Consistent (f)Consistent
(g)Inconsistent(h)Consistent (i)Inconsistent
3. Number of pants = 1; Number of shirts = 0
4. Number of Girls = 7; Number of boys = 4
5. Cost of pencil = 23; Cost of pen = 25
6. Length = 20 m; Width = 16 m
7. (i)3x + 2y 7 = 0
(ii)3x + 3y 12 = 0
(iii)4x + 6y 16 = 0
8. Length = 56 units; Breadth = 100 units
9. Number of students = 16;Number of benches = 5
EXERCISE - 4.2
1. Income of I
st
person = ` 18000; Income of II
nd
person = ` 14000
2. 42 and 24
3. Angles are 54º and 36º
4. Fixed charge = ` 40; Charge per km = ` 18
5.
7
9

384. Answers 3 7 3
Free Distribution by A.P. Government
6. 60 km/h; 40 km/h.
7. 61º and 119º
8. 659 and 723
9. 40 ml and 60 ml
10.` 7200 and ` 4800
EXERCISE - 4.3
1. (i)(4, 5) (ii)
1 1
,
2 4
 

  
(iii)(4, 9)
(iv)(1, 2) (v)(3, 2) (vi)
1 1
,
2 3
 

 
(vii)(3, 2) (viii)(1, 1)
2. (i)Speed of boat = 8 km/h; Speed of streem = 3 km/h
(ii)Speed of train = 60 km/h; Speed of car = 80 km/h
(iii)Number of days by man = 18;Number of days by woman = 36
EXERCISE - 5.1
1. (i)Yes (ii)Yes (iii)No
(iv)Yes (v)Yes (vi)No
(vii)No (viii)Yes
2. (i)2x
2
+ x 528 = 8 (x = Breadth)
(ii)x
2
+ x 306 = 0 (x = Smaller integer)
(iii)x
2
+ 32x 273 = 0 (x = Rohan's Age)
(iv)x
2
 8x + 1280 = 0 (x = Speed of the train)
EXERCISE - 5.2
1. (i)2; 5 (ii)2;
3
2
(iii)2;
5
2

(iv)
1 1
;
4 4
(v)
1 1
;
10 10
(vi)6; 2

385. Mathematic3 7 4
Free Distribution by A.P. Government
(vii)1,
2
3
(viii)1; 3 (ix)7,
8
3
2. 13, 14
3. 17, 18; 17, 18
4. 5 cm, 12 cm
5. Number of articles = 6;Cost of each article = 15
6. 4 m; 10 m
7. Base = 12 cm; Altitide = 8 cm
8. 15 km, 20 km
9. 20 or 40
10.9 kmph
EXERCISE - 5.3
1. (i)
1 33 1 33
,
4 4
   
(ii)
3
2

,
3
2

(iii)
7 71 7 71
,
10 10
  
(iv)1, 5
3. (i)
3 13
2

,
3 13
2

(ii)1, 2
4. 7 years
5. Maths = 12, Einglish = 18(or)Maths = 13,English = 17
6. 120 m; 90 m
7. 18, 12; 18, 12
8. 40 kmph
9. 15 hours,25 hours
10.Speed of the passenger train = 33 kmph
Speed of the express train = 44 kmph
11.18 m;12 m
12.6 seconds
13.13 sides;No

386. Answers 3 7 5
Free Distribution by A.P. Government
EXERCISE - 5.4
1. (i)Read roots do not exist
(ii)Equal roots;
2
3
,
2
3
(iii)Distinct roots;
3 3
2

2. (i) 2 6k  (ii) 6k
3. Yes; 40 m; 20 m
4. No
5. Yes; 20 m; 20 m
EXERCISE - 6.1
1. (i)AP (ii)Not AP (iii)AP (iv)Not AP
2. (i)10, 20, 30, 40 (ii)2, 2, 2, 2
(iii)4, 1, 2, 5 (iv)1,
1
2
, 0,
1
2
(v)1.25, 1.5, 1.75, 2
3. (i)a
1
= 3; d = 2 (ii)a
1
= 5; d = 4
(iii)1
1
3
a;
4
3
d (iv)a
1
= 0.6; d = 1.1
4. (i)Not AP
(ii)AP, next three terms = 4,
9
2
, 5
(iii)AP, next three terms = 9.2, 11.2, 13.2
(iv)AP, next three terms = 6, 10, 14
(v)AP, next three terms = 3 + 4 2, 3 + 5 2, 3 + 6 2
(vi)Not AP
(vii)AP, next three terms = 16, 20, 24

387. Mathematic3 7 6
Free Distribution by A.P. Government
(viii)AP, next three terms =
1
2

,
1
2

,
1
2

(ix)Not AP
(x)AP, next three term = 5a, 6a, 7a
(xi)Not AP
(xii)AP, next three terms =50, 72, 98
(xiii)Not AP
EXERCISE - 6.2
1. (i)a
8
= 28 (ii)d = 2 (iii)a = 46
(iv)n = 10 (v)a
n
= 3.5
2. (i)84 (ii)22
3. (i)a
2
= 14
(ii)a
1
= 18; a
3
= 8
(iii)a
2
=
13
2
; a
3
= 8
(iv)a
2
= 2; a
3
= 0; a
4
= 2; a
5
= 4
(v)a
1
= 53; a
3
= 23; a
4
= 8; a
5
= 7
4. 16
th
term
5. (i)34 (ii)27
6. No 7. 178 8. 5 9. 1
10.100 11.128 12.60 13.13
14.AP = 4, 10, 16, .... 15.158
16.13, 8, 3 17.11 18.13
EXERCISE - 6.3
1. (i)245(ii)180 (iii)555 (iv)
33 13
1
20 20

2. (i)
2093 1
1046
2 2
 (ii)286 (iii)8930
3. (i)440 (ii) 13
7
, 273
3
d S 

388. Answers 3 7 7
Free Distribution by A.P. Government
(iii)a = 4, S
12
= 246 (iv)d = 1, a
10
= 22
(v)x = 5; a
5
= 37 (vi)x = 7;a = 8
(vii)a = 4
4. x = 38; S
38
= 6973
5. 5610
6. x
2
7. (i)525 (ii)465
8. S
1
= 3; S
2
= 4; a
2
= 1; a
3
= 1; a
10
= 15
a
n
= 5 2x
9. 4920 10.160, 140, 120, 100, 80, 60, 40
11.234 12.143 13.16 14.370
EXERCISE - 6.4
1. (i)No (ii)No (iii)Yes
2. (i)4, 12, 36, .... (ii)
5 5
5, , ,...
5 25
(iii)81, 27, 9, .... (iv)
1 1 1
, , , ......
64 32 16
3. (i)Yes; 32, 64, 128 (ii)Yes,
1 1 1
, ,
24 48 96
 
(iii)No (iv)No (v)No
(vi)Yes; 81, 243, 729 (vii)Yes;
2 3 4
1 1 1
, , , ......
x x x
(viii)Yes; 16, 32 2, 128  (ix)Yes; 0.0004, 0.00004, 0.000004
4. 4
EXERCISE - 6.5
1. (i)r
a
= 3;
1
1
3
2
n
n
a

 
  
(ii)r = 3; a
n
= 2(3)
n-1

389. Mathematic3 7 8
Free Distribution by A.P. Government
(iii)r = 3; a
n
= 3(3)
n-1
(iv)r =
2
5
;
1
2
5
5
n
n
a

 
  
2. a
10
= 5
10
; a
n
= 5
n
3. (i)
4
1
3
(ii)
4
4
3

4. (i)5 (ii)12 (iii)7
5. 2
12
6.
9 3
, , 1, ....
4 2
7. 5
EXERCISE - 7.1
1. (i)2 2 (ii)4 2 (iii)5 2 (iv)
2 2
2a b
2. 39
3. Points are not collinear
9. (i)Square (ii)Trepezium (iii)Parallelogram
10.(7, 0) 11.7 or 5
12.3 or 9 13.2 5 units
EXERCISE - 7.2
1. (1, 3) 2.
5
2,
3
 

  
and
7
0,
3
 

  
3. 2 : 7 4. x = 6 ; y = 3
5. (3, 10) 6.
2 20
,
7 7
  

  
7.
3 9
3, , ( 2, 3), 1,
2 2
   
           
8.
13
1,
2
 

 
9. 24 sq. units 10.
5 5
,
5 5
a b a b  

  
11.(i)
2
, 2
3
 

 
(ii)
10 5
,
3 3
 

  
(iii)
2 5
,
3 3
 

  

390. Answers 3 7 9
Free Distribution by A.P. Government
EXERCISE - 7.3
1. (i)
1
2
2
sq. units (ii)32 sq. units(iii)3 sq. units
2. (i)K = 4 (ii)K = 3 (iii)K =
7
3
3. 1 sq. unit ; 1 : 44.
33
2
sq. units 5. 1500 3 sq. units
EXERCISE - 7.4
1. (i)6 (ii)3 (iii)
4b
a
(iv)
a
b

(v)
25
19

(vi)0 (vii)
1
7
(viii)1
EXERCISE - 8.1
4. x = 5 cm and y =
13
2
16
cm or 2.8125 cm
EXERCISE - 8.2
1. (ii)DE = 2.8 cm
2. 8 cm 3. 1.6 m 7. 16 m
EXERCISE - 8.3
3. 1: 4 4.
2 1
1

6. 96 cm
2
8. 3.5 cm
EXERCISE - 8.4
8. 6 7 m 9. 13 m 12.1: 2
EXERCISE - 9.1
1. (i)One (ii)Secant of a circle(iii)Two
(iv)Point of contact(v)Infinite
2. PQ = 13 cm 4. 306 cm

391. Mathematic3 8 0
Free Distribution by A.P. Government
EXERCISE - 9.2
1. (i)d (ii)a (iii)b (iv)a (v)c
2. 8 cm 4. AB = 15 cm, AC = 13 cm
5. 8 cm each6. 2 5cm 9. Two
EXERCISE - 9.3
1. (i)28.5 cm
2
(ii)285.5 cm
2
2. 88.368 cm
2
3. 1254.96 cm
2
4. 57 cm
2
5. 10.5 cm
2
6. 9.625 cm
2
7. 102.67 cm
2
8. 57 cm
2
EXERCISE - 10.1
1. 5500 cm
2
2. 124800 cm
2
(12.48 m
2
) 3. 264 c.c.
4. 1 : 2 5. 4772 7. 29645 cm
3
8. 188.57 m
2
9. 37 cm
EXERCISE - 10.2
1. 103.71 cm
2
2. 1156.57 cm
2
3. 220 mm
2
4. 160 cm
2
5. ` 765.6 6. 4 : 4 : 5
7.
2
5
2
a
 
  
sq. units 8. 374 cm
2
EXERCISE - 10.3
1. 693 kg 2. Height of cone = 22.05 cm;Surface area of toy = 793 cm
2
3. 88.83 cm
3
4. 616 cm
3
5. 309.57 cm
3
6. 150 7. 523.9 cm
3
EXERCISE - 10.4
1. 2.74 cm 2. 12 cm 3. 0.714 m (71.4 cm)
4. 5 m 5. 10 6. 57
7. 100 8. 224

392. Answers 3 8 1
Free Distribution by A.P. Government
EXERCISE - 11.1
1. sin A =
15
17
; cos A =
18
17
; tan A =
15
8
2.
527
168
3. cos =
49
25
; tan=
24
49
4. sin A =
5
13
; tan A =
5
12
5. sin A =
4
5
; cos A =
3
5
7. (i)
47
62
(ii)
111 8
7

8. (i)1 (ii)0
EXERCISE - 11.2
1. (i) 2 (ii)
3
4 2
(iii)1
(iv)
1
3

(v)1
2. (i)c (ii)d (iii)b
3. 1 4. Yes
5. QR = 6 3cm; PR = 12 cm
6. YXZ = 60º; YXZ = 30º 7. It is true
EXERCISE - 11.3
1. (i)1 (ii)0 (iii)0
(iv)1 (v)1
3. A = 24º 6. cos 15º + sin 25º
EXERCISE - 11.4
1. (i)2 (ii)2 (iii)1
6. 1 8. 1 9.
1
p

393. Mathematic3 8 2
Free Distribution by A.P. Government
EXERCISE - 12.1
1. 15 m 2. 63 m 3. 4 m
4. 60º 5. 11.55 m 6. 43 m
7. 4.1568 m 8. 300 m 9. 15 m 10.12.99 cm
2
EXERCISE - 12.2
1. Height of the tower = 53m; Width of the road = 5 m
2. 32.908 m 3. 1.464 m 4. 19.124 m
5. 7.608 m 6. 10 m 7. 51.96 feets;30 feets
9. 200 m/sec.10.24 m
EXERCISE - 13.1
1. (i)1 (ii)0, Impossible event(iii)1, Sure event
(iv)1 (v)0, 1
2. (i)No (ii)No (iii)Yes(iv)Yes
3. 0.95 4. (i)0 (ii)1
5. 0.008 6. (i)
1
2
(ii)
1
2
(iii)
1
2
EXERCISE - 13.2
1. (i)
3
8
(ii)
5
8
2. (i)
5
17
(ii)
4
17
(iii)
13
17
3. (i)
5
9
(ii)
17
18
4.
5
13
5. 0.35
6. (i)
1
8
(ii)
1
2
(iii)
3
4
(iv)1

394. Answers 3 8 3
Free Distribution by A.P. Government
7. (i)
1
26
(ii)
1
13
(iii)
1
26
(iv)
1
52
(v)
1
13
(vi)
1
52
8.
3
10
9.
4
15
10.(i)
1
5
(ii)a.
1
4
b.
1
4
11.
11
12
12.(i)
1
5
(ii)
15
19
13.(i)
9
10
(ii)
1
10
(iii)
1
5
14.
11
21
15.(i)
31
36
(ii)
5
36
16.
Sum on 2 dice 2 3 4 5 6 7 8 9 10 11 12
Probability
1
36

1
18

1
12

1
9

5
36

1
6

5
36

1
9

1
12

1
18

1
36
17.(i)
1
2
(ii)
1
2
EXERCISE - 14.1
1. 8.1 plants. We have used direct method because numerical values of x
i
and f
i
are small.
2. ` 145.20 3. f = 20 4. 75.9
5. 22.31 6. ` 211 7. 0.099 ppm
8. 49 days 9. 69.43%
EXERCISE - 14.2
1. Mode = 36.8 years, Mean = 35.37 years, Maximum number of patients admitted in the
hospital are of the age 36.8 years (approx.), while on an average the age of a patient
admitted to the hospital is 35.37 years.

395. Mathematic3 8 4
Free Distribution by A.P. Government
2. 65.625 hours
3. Modal monthly expenditure = ` 1847.83, Mean monthly expenditure = ` 2662.5.
4. Mode : 30.6, Mean = 29.2. Most states/U.T. have a student teacher ratio of 30.6 and
on an average, this ratio is 29.2.
5. Mode = 4608.7 runs.
6. Mode = 44.7 cars
EXERCISE - 14.3
1. Median = 137 units, Mean = 137.05 units, Mode = 135.76 units.
The three measures are approximately the same in this case.
2. x = 8, y = 7
3. Median age = 35.76 years
4. Median length = 146.75 mm
5. Median life = 3406.98 hours
6. Median = 8.05, Mean = 8.32, Modal size = 7.88
7. Median weight = 56.67 kg
EXERCISE - 14.4
1. Daily income (in `)Cumulative frequency
Less than 120 12
Less than 140 26
Less than 160 34
Less than 180 40
Less than 200 50
2. Draw the ogive by plotting the points : (38, 0), (40, 3), (42, 5), (44, 9), (46, 14),
(48, 28), (50, 32) and (52, 35). Here
2
n
=17.5. Locate the point on the ogive whose
ordinate is 17.5. The x-coordinate of this point will be the median.
Draw ogive by plotting the points :
(120, 12), (140, 26), (160, 34),
(180, 40) and (200, 50)

396. Answers 3 8 5
Free Distribution by A.P. Government
3. Production yield (kg/ha) Cumulative frequency
More than or equal to 50 100
More than or equal to 55 98
More than or equal to 60 90
More than or equal to 65 78
More than or equal to 70 54
More than or equal to 75 16
Now, draw the ogive by plotting the points : (50, 100), (55, 98), (60, 90), (65, 78),
(70, 54) and (75, 16).